Which statement about thermal energy transfers is correct?
Show complete worked solution
Correct answer: B
Recall the defining physical principle for thermal energy transfers.
A larger specific heat capacity means more energy is required for the same mass and temperature rise.
This matches option B.
QUESTION 2[1 mark]
Multiple choiceFoundationCalculatorAbout 1 min
Which equation is a valid starting point for analysing thermal energy transfers?
Show complete worked solution
Correct answer: B
Identify the quantities involved.
The relevant syllabus relationship is $$Q=mc\Delta T$$
This is option B.
QUESTION 3[3 marks]
Extended responseStandardCalculatorAbout 2.4 min
Explain the physical meaning of the relationship $$Q=mc\Delta T$$ in the context of thermal energy transfers.
Show complete worked solution
Identify the dependent quantity: energy transferred.
Identify the changing quantity: temperature change, while the other quantities in the equation are controlled.
The equation shows that energy transferred is proportional to temperature change. State this with the physical conditions made explicit.
Final answer: thermal energy transferred is proportional to temperature change for fixed mass and material
QUESTION 4[3 marks]
Data-basedStandardCalculatorAbout 2.4 min
The graph shows how energy transferred varies with temperature change. State the trend and explain whether it is consistent with $$Q=mc\Delta T$$.
Show complete worked solution
Read the graph shape rather than a single point.
The plotted trend represents energy transferred is proportional to temperature change.
This is consistent with $$Q=mc\Delta T$$ when the remaining quantities are constant.
Final answer: energy transferred is proportional to temperature change
QUESTION 5[5 marks]
Extended responseAppliedCalculatorAbout 4 min
Design a practical investigation of the relationship used in thermal energy transfers. Identify the independent variable, dependent variable and two important controls.
Show complete worked solution
Choose temperature change as the independent variable and energy transferred as the dependent variable.
Measure mass and temperature with insulation, supply measured electrical energy, and use the gradient to determine specific heat capacity.
Keep the other quantities in $$Q=mc\Delta T$$ constant and repeat readings.
Plot a graph that linearizes the predicted relationship and include uncertainty bars where possible.
Final answer: Measure mass and temperature with insulation, supply measured electrical energy, and use the gradient to determine specific heat capacity.
QUESTION 6[4 marks]
Multi-partAppliedCalculatorAbout 3.2 min
For a system obeying the relevant proportional relationship in thermal energy transfers, temperature change doubles while all required control variables remain constant. Determine the factor change in energy transferred and justify it.
Show complete worked solution
From $$Q=mc\Delta T$$ isolate the proportional dependence: energy transferred is proportional to temperature change.
Form a ratio for the new and old situations.
Substituting $x_2=2x_1$ gives $$\frac{y_2}{y_1}=2$$ Therefore energy transferred changes by a factor of $2$.
Final answer: 2
QUESTION 7[4 marks]
Data-basedAppliedCalculatorAbout 3.2 min
A normalized data set gives $x_1=1.00$ and $y_1=1.00$. When $x_2=2.00$, the measured value is $y_2=2$. Assess whether the data support the expected model for thermal energy transfers.
Show complete worked solution
The model predicts energy transferred is proportional to temperature change.
For $x_2/x_1=2$, the predicted ratio is $y_2/y_1=2$.
The measured normalized ratio is also $2$. The data therefore support the model within the given precision; a stronger conclusion would require uncertainties and more points.
Final answer: The data support the model within the stated precision.
QUESTION 8[1 mark]
Multiple choiceStandardCalculatorAbout 1 min
Which experimental practice would most improve the reliability of a test of thermal energy transfers?
Show complete worked solution
Correct answer: B
Reliability concerns the consistency of repeated measurements.
Repeats reveal random variation and allow a mean to be calculated.
Monitoring controls prevents a second variable from creating a false trend. Option B is correct.
QUESTION 9[5 marks]
Extended responseAdvancedCalculatorAbout 4 min
A student says: “A larger specific heat capacity means more energy is required for the same mass and temperature rise. Therefore the result is exact and no assumptions are involved.” Evaluate this claim.
Show complete worked solution
The physical principle is valid: A larger specific heat capacity means more energy is required for the same mass and temperature rise.
The conclusion that a measurement is exact does not follow. Models assume stated conditions, instruments have finite resolution, and uncontrolled effects may be present.
A valid evaluation reports uncertainty, tests controls and states the range over which $$Q=mc\Delta T$$ applies.
Final answer: The principle is valid, but any application must state assumptions and account for measurement uncertainty.
QUESTION 10[6 marks]
Multi-partAdvancedCalculatorAbout 4.8 min
Construct a complete analysis strategy for an unfamiliar problem involving thermal energy transfers: state the governing equation, describe the diagram or graph you would draw, and give one independent check on the result.
Show complete worked solution
Begin with the governing relationship $$Q=mc\Delta T$$ and define every symbol.
Draw the relevant system boundary or axes and label temperature change and energy transferred.
Substitute only after converting to SI units; keep extra digits until the end.
Check dimensional consistency and test a limiting case.
State the final result with units, direction where relevant, and appropriate significant figures.
Final answer: Use $$Q=mc\Delta T$$, represent the variables clearly, and check units, limiting behaviour or energy/momentum consistency.
QUESTION 11[1 mark]
Multiple choiceFoundationCalculatorAbout 2 min
A sample warms from $18\,^{\circ}\mathrm{C}$ to $53\,^{\circ}\mathrm{C}$. What is the temperature change expressed in kelvin?
Show complete worked solution
Correct answer: A
Celsius and kelvin scales have equal-size intervals.
$$\Delta T=53-18=35$$ A temperature change of $35\,^{\circ}\mathrm{C}$ equals $35\,\mathrm{K}$. Option A is correct.
QUESTION 12[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A solid cube of side $L$ absorbs energy $Q$ and its temperature rises by $\Delta T$. A second cube of the same material has side $3L$ and absorbs $6Q$. What is its temperature rise?
Show complete worked solution
Correct answer: A
For the same material, mass is proportional to volume. The second cube has mass $3^3=27$ times larger.
Since $Q=mc\Delta T$, $$\frac{\Delta T_2}{\Delta T_1}=\frac{Q_2/Q_1}{m_2/m_1}=\frac{6}{27}=\frac29$$ Therefore option A is correct.
QUESTION 13[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
Three samples of the same liquid are mixed in an insulated container: $3M$ at $70^\circ\mathrm{C}$, $M$ at $40^\circ\mathrm{C}$, and $2M$ at $10^\circ\mathrm{C}$. What is the final temperature?
Show complete worked solution
Correct answer: C
With the same liquid and no heat loss, use the mass-weighted mean.
$$T_f=\frac{3M(70)+M(40)+2M(10)}{6M}$$
$$T_f=\frac{270}{6}=45^\circ\mathrm{C}$$ Option C is correct.
QUESTION 14[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
Blocks X and Y reach thermal equilibrium in an insulated enclosure. X has mass $2m$, specific heat capacity $c$, and warms from $20^\circ\mathrm C$ to $50^\circ\mathrm C$. Y has specific heat capacity $3c$ and cools from $80^\circ\mathrm C$ to $50^\circ\mathrm C$. What is the mass of Y?
Show complete worked solution
Correct answer: A
Heat gained by X is $Q_X=(2m)c(50-20)=60mc$.
Heat lost by Y is $Q_Y=m_Y(3c)(80-50)=90m_Yc$.
Equating energies gives $60mc=90m_Yc$, so $m_Y=2m/3$ and A is correct.
QUESTION 15[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A liquid vaporizes at constant temperature. Which quantities are greater in the gas phase? I. Total intermolecular potential energy. II. Root-mean-square molecular speed. III. Average molecular separation.
Show complete worked solution
Correct answer: B
Latent heat separates molecules against attractions, increasing intermolecular potential energy.
At unchanged temperature, average kinetic energy and rms speed remain unchanged for the same molecules.
Gas molecules are much farther apart, so I and III only are greater; B is correct.
QUESTION 16[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
An insulated container holds liquid of mass $3M$ that warms by $8^\circ\mathrm C$ when a metal sample of mass $M$ cools by $48^\circ\mathrm C$. What is $c_{liquid}/c_{metal}$?
Show complete worked solution
Correct answer: C
Energy gained by the liquid equals energy lost by the metal.
Thus $(3M)c_l(8)=M c_m(48)$.
Cancelling common factors gives $c_l/c_m=48/24=2$, so C is correct.
QUESTION 17[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
The same liquid sample needs $10\,\mathrm{kJ}$ to warm by $5.0\,\mathrm K$ and $200\,\mathrm{kJ}$ to vaporize at constant temperature. What is $L_v/c$, where $L_v$ is specific latent heat and c is specific heat capacity?
Show complete worked solution
Correct answer: C
For heating, $10=mc(5.0)$, so $mc=2.0\,\mathrm{kJ\,K^{-1}}$.
For vaporization, $200=mL_v$.
Dividing gives $L_v/c=200/(mc)=200/2.0=100\,\mathrm K$, so C is correct.
QUESTION 18[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Equal masses of a substance are heated at the same constant power. Its temperature-time graph is less steep in the liquid phase than in the solid phase. What follows?
Show complete worked solution
Correct answer: A
Away from phase changes, slope is $dT/dt=P/(mc)$.
For equal mass and power, a smaller slope means a larger c.
The liquid segment is less steep, so liquid specific heat capacity is larger and A is correct.
QUESTION 19[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
Water of mass m cools from $290\,\mathrm K$ to $273\,\mathrm K$ when ice at its melting point is added. Water specific heat is c and ice latent heat of fusion is L. What ice mass melts?
Show complete worked solution
Correct answer: B
Cooling water releases $Q=mc(290-273)=17mc$.
Ice at melting point uses energy $m_iL$ to melt.
Equating gives $m_i=17mc/L$, so B is correct.
QUESTION 20[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Water is boiling at constant pressure while supplied with power P. The input power is increased. What happens to its temperature and vaporization rate?
Show complete worked solution
Correct answer: C
During boiling at fixed pressure, added energy supplies latent heat rather than raising temperature.
Mass vaporization rate is $P/L_v$.
Greater P increases vaporization rate while temperature stays at the boiling point. Therefore option C is correct.
QUESTION 21[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A tube containing metal spheres is inverted N times. Each sphere falls through height h during every inversion. Neglect losses outside the spheres. What is the metal’s specific heat capacity if its temperature rises by $\Delta T$?
Show complete worked solution
Correct answer: D
Each sphere loses gravitational energy $mgh$ per inversion, so total per sphere is $Nmgh$.
The gained internal energy per sphere is $mc\Delta T$.
Equating and cancelling m gives $c=Nhg/\Delta T$. Therefore option D is correct.
QUESTION 22[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A $1.5\,\mathrm{kg}$ liquid of specific heat capacity $3000\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is removed from a $450\,\mathrm W$ heater at equilibrium temperature. What is its initial cooling rate?
Show complete worked solution
Correct answer: A
At equilibrium on the heater, heat loss to the surroundings equals 450 W.
Immediately after removal, the initial loss rate is still approximately 450 W, so $P=mc|dT/dt|$.
$|dT/dt|=450/[1.5(3000)]=0.10\,\mathrm{K\,s^{-1}}$. Therefore option A is correct.
QUESTION 23[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A solid sublimes to a gas at constant temperature. What happens to internal energy and intermolecular potential energy?
Show complete worked solution
Correct answer: D
Constant temperature means average molecular kinetic energy stays approximately unchanged.
Energy is required to separate molecules against attractive forces, increasing potential energy.
Internal energy includes this potential energy, so both increase. Therefore option D is correct.
QUESTION 24[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
A liquid of mass m and specific heat capacity c cools at rate $|dT/dt|=k$. At what rate does it lose thermal energy?
Show complete worked solution
Correct answer: C
Thermal energy change is $dQ=mc\,dT$.
Dividing by time gives $|dQ/dt|=mc|dT/dt|$.
Substituting k gives the rate $mck$. Therefore option C is correct.
QUESTION 25[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A pure solid has just reached its melting point and receives energy at constant rate. How does temperature vary with supplied energy while melting?
Show complete worked solution
Correct answer: C
During a phase change, supplied energy increases intermolecular potential energy.
The average kinetic energy and therefore temperature stay constant until all solid melts.
The result is It remains constant until melting is complete. Therefore option C is correct.
QUESTION 26[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
What is the unit of $c/L_v$, where c is specific heat capacity and $L_v$ is specific latent heat?
Show complete worked solution
Correct answer: C
$[c]=\mathrm{J\,kg^{-1}\,K^{-1}}$ and $[L_v]=\mathrm{J\,kg^{-1}}$.
Dividing cancels J kg$^{-1}$ and leaves K$^{-1}$.
The result is $\mathrm{K^{-1}}$ Therefore option C is correct.
QUESTION 27[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A liquid heated at constant power P has temperature-time gradient k and specific heat capacity c. What is its mass?
Show complete worked solution
Correct answer: C
Heating power is $P=mc(dT/dt)$.
With gradient k, rearrange to $m=P/(ck)$.
The result is $P/(ck)$ Therefore option C is correct.
QUESTION 28[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In a satellite measurement at constant temperature, a student says the thermal result is exact because it matches $Q=mc\Delta T$. Which evaluation is strongest?
Show complete worked solution
Correct answer: A
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $Q=mc\Delta T$. This corresponds to option A.
QUESTION 29[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which approach gives a dimensionally consistent calculation for an unknown in $Q=mc\Delta T$ during a field-mapping exercise with negligible losses?
Show complete worked solution
Correct answer: C
First isolate the requested symbol algebraically.
Convert every measured quantity to coherent SI units, substitute, then round only the final result.
The supported conclusion is Rearrange $Q=mc\Delta T$ before substituting values, and keep all quantities in SI units. This corresponds to option C.
QUESTION 30[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which approach gives a dimensionally consistent calculation for an unknown in $Q=mc\Delta T$ during a field-mapping exercise with repeated readings?
Show complete worked solution
Correct answer: C
First isolate the requested symbol algebraically.
Convert every measured quantity to coherent SI units, substitute, then round only the final result.
The supported conclusion is Rearrange $Q=mc\Delta T$ before substituting values, and keep all quantities in SI units. This corresponds to option C.
QUESTION 31[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
During a particle-detector experiment with SI data, temperature change increases by $65\%$. What percentage change in thermal energy follows from $Q=mc\Delta T$?
Show complete worked solution
Correct answer: B
Convert the percentage change to the multiplier $x_2/x_1=1.65$.
Use $y_2/y_1=(1.65)^{1}=1.65$ and convert the ratio back to a percentage change.
The supported conclusion is $65\%$ increase This corresponds to option B.
QUESTION 32[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a thermal-control system using two matched systems, temperature change is multiplied by $2$ and then by $0.71$. By what overall factor does thermal energy change according to $Q=mc\Delta T$?
Show complete worked solution
Correct answer: C
The total multiplier of temperature change is $2\times0.71=1.42$.
Raise this multiplier to the power $1$, giving $1.42$.
The supported conclusion is $1.42$ This corresponds to option C.
QUESTION 33[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which condition is required when using $Q=mc\Delta T$ to compare two measurements of thermal energy in a laboratory calibration with an idealized component?
Show complete worked solution
Correct answer: D
Identify which variables appear in the complete physical relation.
A one-variable proportional comparison is valid only when the remaining variables are controlled.
The supported conclusion is The prediction follows only if the quantities omitted from $Q=mc\Delta T$ remain constant. This corresponds to option D.
QUESTION 34[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
In a field-mapping exercise using a reference sample, by what factor must temperature change change for thermal energy to change by a factor of $1.75$?
Show complete worked solution
Correct answer: B
Use the ratio equation $1.75=(x_2/x_1)^{1}$.
Take the power $1/1$ to obtain $x_2/x_1=1.75$.
The supported conclusion is $1.75$ This corresponds to option B.
QUESTION 35[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from a circuit-design trial at constant temperature agree with the thermal prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: A
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $Q=mc\Delta T$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option A.
QUESTION 36[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
Which graph would best test the predicted dependence in an engineering prototype over a controlled range involving thermal?
Show complete worked solution
Correct answer: C
Rearrange $Q=mc\Delta T$ into a linear form.
Choose axes that predict a straight line, then assess gradient, intercept and uncertainty bars.
The supported conclusion is Plot thermal energy against a transformed temperature change chosen to make $Q=mc\Delta T$ linear. This corresponds to option C.
QUESTION 37[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In an optical-instrument test using a computer interface, a student says the thermal result is exact because it matches $Q=mc\Delta T$. Which evaluation is strongest?
Show complete worked solution
Correct answer: D
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $Q=mc\Delta T$. This corresponds to option D.
QUESTION 38[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
What is the strongest experimental test of the mathematical form of the thermal model in an environmental monitor under steady conditions?
Show complete worked solution
Correct answer: B
A model test requires a range of the independent variable and repeated measurements.
Linearize the proposed relation in advance and judge agreement using gradient, intercept and uncertainties.
The supported conclusion is Use several values of temperature change and test whether the transformed graph predicted by $Q=mc\Delta T$ is linear. This corresponds to option B.
QUESTION 39[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In an optical-instrument test using two matched systems, a student says the thermal result is exact because it matches $Q=mc\Delta T$. Which evaluation is strongest?
Show complete worked solution
Correct answer: D
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $Q=mc\Delta T$. This corresponds to option D.
QUESTION 40[8 marks]
Data-basedFoundationCalculatorAbout 10 min
The graph shows an idealized heating curve for a $0.33\,\mathrm{kg}$ substance heated at $540\,\mathrm{W}$. In its first phase, $c=2600\,\mathrm{J\,kg^{-1}K^{-1}}$; the first temperature rise is $18\,\mathrm{K}$ and the latent heat for the first plateau is $2.20e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the time for the first temperature rise.[3]
b. Calculate the duration of the first plateau.[3]
c. Explain the sloping and horizontal sections of the graph.[2]
On a slope, energy increases mean molecular kinetic energy, so temperature rises. On a plateau, energy changes molecular separation and potential energy, so temperature stays constant.
Final answer: Slopes show rising mean kinetic energy; plateaus show increasing potential energy during a phase change.
QUESTION 41[6 marks]
Multi-partAdvancedCalculatorAbout 7.5 min
A wool jacket uses trapped air pockets to control thermal-energy transfer.
a. Identify the principal transfer mechanism affected.[1]
b. Explain the effect using a particle or radiation model.[3]
c. Suggest one measurement that could test the effectiveness of the design.[2]
Show complete worked solution
(a)
The design feature is chosen to reduce conduction and convection. The relevant mechanism is therefore conduction and convection.
Final answer: Conduction and convection.
(b)
Trapped air pockets reduce conduction and convection. The explanation follows from particle collisions and bulk fluid motion for conduction/convection, or absorption and emission properties for radiation.
Final answer: Reduce conduction and convection.
(c)
Use identical systems with the same initial temperature difference. Change only the named feature, record temperature at equal time intervals, and compare the cooling or heating rates.
Final answer: Measure temperature change over equal times with and without the feature.
QUESTION 42[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A temperature sensor records $18.0\,\mathrm{^\circ C}$. It later records a temperature $32.0\,\mathrm{K}$ higher.
a. Convert the initial temperature to kelvin.[2]
b. Determine the final temperature in degrees Celsius.[2]
c. State why kelvin must be used in molecular-energy equations.[2]
Show complete worked solution
(a)
$$T=18.0+273.15=\boxed{291.15\,\mathrm{K}}$$
Final answer: $291.15\,\mathrm{K}$
(b)
A temperature interval has the same numerical value in kelvin and degrees Celsius.
$$\theta_f=18.0+32.0=\boxed{50.0\,\mathrm{^\circ C}}$$
Final answer: $50.0\,\mathrm{^\circ C}$
(c)
Quantities such as average molecular kinetic energy are proportional to absolute temperature. Celsius has an arbitrary zero, so substituting Celsius would not preserve that proportionality.
Final answer: Kelvin is an absolute thermodynamic scale with zero at minimum thermal energy.
QUESTION 43[7 marks]
Multi-partAdvancedCalculatorAbout 8.8 min
A star has luminosity $5.60e+29\,\mathrm{W}$ and is $4.60e+18\,\mathrm{m}$ from Earth.
a. Calculate its apparent brightness at Earth.[3]
b. State why luminosity and apparent brightness are different quantities.[2]
c. Determine the distance at which the apparent brightness would be one quarter as large.[2]
A cylindrical heating element of radius $1.00\,\mathrm{mm}$ and length $0.89\,\mathrm{m}$ is at $835\,\mathrm{K}$. Its emissivity is $0.90$; neglect radiation from its ends.
a. Calculate its radiating surface area.[2]
b. Calculate the emitted radiation power.[3]
c. Determine the factor change in power if the absolute temperature rises by $10\%$.[2]
$\rho_{aluminium}=2700\,\mathrm{kg\,m^{-3}}$ and $\rho_{water}\approx1000\,\mathrm{kg\,m^{-3}}$.
The sample is denser, so its weight exceeds the maximum buoyant force before full support and it sinks.
Final answer: It sinks because its density exceeds that of water.
QUESTION 46[8 marks]
Data-basedFoundationCalculatorAbout 10 min
The graph shows an idealized heating curve for a $0.37\,\mathrm{kg}$ substance heated at $600\,\mathrm{W}$. In its first phase, $c=2850\,\mathrm{J\,kg^{-1}K^{-1}}$; the first temperature rise is $18\,\mathrm{K}$ and the latent heat for the first plateau is $2.40e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the time for the first temperature rise.[3]
b. Calculate the duration of the first plateau.[3]
c. Explain the sloping and horizontal sections of the graph.[2]
On a slope, energy increases mean molecular kinetic energy, so temperature rises. On a plateau, energy changes molecular separation and potential energy, so temperature stays constant.
Final answer: Slopes show rising mean kinetic energy; plateaus show increasing potential energy during a phase change.
QUESTION 47[7 marks]
Multi-partAdvancedCalculatorAbout 8.8 min
A star has luminosity $2.80e+29\,\mathrm{W}$ and is $3.00e+18\,\mathrm{m}$ from Earth.
a. Calculate its apparent brightness at Earth.[3]
b. State why luminosity and apparent brightness are different quantities.[2]
c. Determine the distance at which the apparent brightness would be one quarter as large.[2]
A $0.24\,\mathrm{kg}$ hot sample with specific heat capacity $450\,\mathrm{J\,kg^{-1}K^{-1}}$ at $86\,\mathrm{^\circ C}$ is placed in $0.40\,\mathrm{kg}$ of water at $21\,\mathrm{^\circ C}$. Heat exchange with the cup and surroundings is negligible.
a. Write the energy-balance equation for the final temperature $T_f$.[2]
b. Calculate the equilibrium temperature.[4]
c. State how including the cup would affect the calculated final temperature.[2]
Show complete worked solution
(a)
Energy lost by the hot sample equals energy gained by the water:
$$m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$$
Final answer: $m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$
(b)
$$(0.24)(450)(86-T_f)=(0.40)(4180)(T_f-21)$$
Expanding and collecting $T_f$ terms gives
$$T_f=\boxed{24.9\,\mathrm{^\circ C}}$$
Final answer: $24.9\,\mathrm{^\circ C}$
(c)
The cup also gains energy. The hot sample must warm both the water and the cup, so the equilibrium temperature lies below the value calculated when the cup is ignored.
Final answer: The final temperature would be lower if the cup initially has the water temperature.
QUESTION 49[7 marks]
Multi-partAppliedCalculatorAbout 8.8 min
A cylindrical heating element of radius $0.90\,\mathrm{mm}$ and length $0.81\,\mathrm{m}$ is at $790\,\mathrm{K}$. Its emissivity is $0.86$; neglect radiation from its ends.
a. Calculate its radiating surface area.[2]
b. Calculate the emitted radiation power.[3]
c. Determine the factor change in power if the absolute temperature rises by $10\%$.[2]
A cavity wall uses trapped air to control thermal-energy transfer.
a. Identify the principal transfer mechanism affected.[1]
b. Explain the effect using a particle or radiation model.[3]
c. Suggest one measurement that could test the effectiveness of the design.[2]
Show complete worked solution
(a)
The design feature is chosen to reduce conduction and convection. The relevant mechanism is therefore conduction and convection.
Final answer: Conduction and convection.
(b)
Trapped air reduce conduction and convection. The explanation follows from particle collisions and bulk fluid motion for conduction/convection, or absorption and emission properties for radiation.
Final answer: Reduce conduction and convection.
(c)
Use identical systems with the same initial temperature difference. Change only the named feature, record temperature at equal time intervals, and compare the cooling or heating rates.
Final answer: Measure temperature change over equal times with and without the feature.
QUESTION 51[8 marks]
Multi-partStandardCalculatorAbout 10 min
A room has an opaque wall area of $14\,\mathrm{m^2}$ and a single-glazed window area of $2.2\,\mathrm{m^2}$. The wall has $k=0.72\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.22\,\mathrm{m}$; the glass has $k=1.0\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.006\,\mathrm{m}$. The temperature difference is $9\,\mathrm{K}$.
a. Determine the power conducted through the opaque wall.[3]
b. Determine the power conducted through the window.[3]
c. Identify which part needs improved insulation and justify quantitatively.[2]
$\bar E_k\propto T$, so doubling $T$ doubles $\bar E_k$.
Final answer: It doubles.
QUESTION 53[8 marks]
Multi-partStandardCalculatorAbout 10 min
A $0.21\,\mathrm{kg}$ hot sample with specific heat capacity $900\,\mathrm{J\,kg^{-1}K^{-1}}$ at $89\,\mathrm{^\circ C}$ is placed in $0.36\,\mathrm{kg}$ of water at $20\,\mathrm{^\circ C}$. Heat exchange with the cup and surroundings is negligible.
a. Write the energy-balance equation for the final temperature $T_f$.[2]
b. Calculate the equilibrium temperature.[4]
c. State how including the cup would affect the calculated final temperature.[2]
Show complete worked solution
(a)
Energy lost by the hot sample equals energy gained by the water:
$$m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$$
Final answer: $m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$
(b)
$$(0.21)(900)(89-T_f)=(0.36)(4180)(T_f-20)$$
Expanding and collecting $T_f$ terms gives
$$T_f=\boxed{27.7\,\mathrm{^\circ C}}$$
Final answer: $27.7\,\mathrm{^\circ C}$
(c)
The cup also gains energy. The hot sample must warm both the water and the cup, so the equilibrium temperature lies below the value calculated when the cup is ignored.
Final answer: The final temperature would be lower if the cup initially has the water temperature.
QUESTION 54[7 marks]
Multi-partAppliedCalculatorAbout 8.8 min
A cylindrical heating element of radius $1.10\,\mathrm{mm}$ and length $0.97\,\mathrm{m}$ is at $880\,\mathrm{K}$. Its emissivity is $0.94$; neglect radiation from its ends.
a. Calculate its radiating surface area.[2]
b. Calculate the emitted radiation power.[3]
c. Determine the factor change in power if the absolute temperature rises by $10\%$.[2]
A solar collector uses a matt-black plate to control thermal-energy transfer.
a. Identify the principal transfer mechanism affected.[1]
b. Explain the effect using a particle or radiation model.[3]
c. Suggest one measurement that could test the effectiveness of the design.[2]
Show complete worked solution
(a)
The design feature is chosen to increase absorption of radiation. The relevant mechanism is therefore radiation.
Final answer: Radiation.
(b)
A matt-black plate increase absorption of radiation. The explanation follows from particle collisions and bulk fluid motion for conduction/convection, or absorption and emission properties for radiation.
Final answer: Increase absorption of radiation.
(c)
Use identical systems with the same initial temperature difference. Change only the named feature, record temperature at equal time intervals, and compare the cooling or heating rates.
Final answer: Measure temperature change over equal times with and without the feature.
QUESTION 56[6 marks]
Multi-partFoundationCalculatorAbout 7.5 min
A flat insulating panel has area $0.94\,\mathrm{m^2}$, thickness $0.024\,\mathrm{m}$ and thermal conductivity $0.80\,\mathrm{W\,m^{-1}\,K^{-1}}$. Its faces differ in temperature by $28\,\mathrm{K}$.
a. Calculate the steady rate of thermal-energy transfer.[3]
b. Calculate the energy transferred in $12.0$ minutes.[2]
c. State the effect of doubling the thickness while other quantities remain fixed.[1]
The graph shows an idealized heating curve for a $0.25\,\mathrm{kg}$ substance heated at $420\,\mathrm{W}$. In its first phase, $c=2100\,\mathrm{J\,kg^{-1}K^{-1}}$; the first temperature rise is $18\,\mathrm{K}$ and the latent heat for the first plateau is $1.80e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the time for the first temperature rise.[3]
b. Calculate the duration of the first plateau.[3]
c. Explain the sloping and horizontal sections of the graph.[2]
On a slope, energy increases mean molecular kinetic energy, so temperature rises. On a plateau, energy changes molecular separation and potential energy, so temperature stays constant.
Final answer: Slopes show rising mean kinetic energy; plateaus show increasing potential energy during a phase change.
QUESTION 58[8 marks]
Multi-partStandardCalculatorAbout 10 min
A room has an opaque wall area of $17\,\mathrm{m^2}$ and a single-glazed window area of $2.8\,\mathrm{m^2}$. The wall has $k=0.72\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.25\,\mathrm{m}$; the glass has $k=1.0\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.009\,\mathrm{m}$. The temperature difference is $15\,\mathrm{K}$.
a. Determine the power conducted through the opaque wall.[3]
b. Determine the power conducted through the window.[3]
c. Identify which part needs improved insulation and justify quantitatively.[2]
$\bar E_k\propto T$, so doubling $T$ doubles $\bar E_k$.
Final answer: It doubles.
QUESTION 60[8 marks]
Multi-partStandardCalculatorAbout 10 min
A room has an opaque wall area of $15\,\mathrm{m^2}$ and a single-glazed window area of $2.4\,\mathrm{m^2}$. The wall has $k=0.72\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.23\,\mathrm{m}$; the glass has $k=1.0\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.007\,\mathrm{m}$. The temperature difference is $11\,\mathrm{K}$.
a. Determine the power conducted through the opaque wall.[3]
b. Determine the power conducted through the window.[3]
c. Identify which part needs improved insulation and justify quantitatively.[2]
$$\frac{P_g}{P_w}=\frac{3771}{517}=7.30$$
The larger transfer rate identifies the dominant path.
Final answer: The window; it has the larger heat-transfer rate.
QUESTION 61[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A heater supplies constant power $1000\,\mathrm{W}$ during the melting of a metal. The mass changing phase is $0.064\,\mathrm{kg}$ and the specific latent heat is $2.050e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the energy required for the phase change.[2]
b. Determine the minimum time required.[2]
c. Explain why the temperature remains approximately constant during the phase change.[2]
The supplied energy separates or rearranges particles against intermolecular forces. Mean kinetic energy, and therefore temperature, remains approximately constant until the phase change is complete.
Final answer: Energy increases intermolecular potential energy rather than mean molecular kinetic energy.
QUESTION 62[8 marks]
Multi-partStandardCalculatorAbout 10 min
A $0.18\,\mathrm{kg}$ hot sample with specific heat capacity $385\,\mathrm{J\,kg^{-1}K^{-1}}$ at $92\,\mathrm{^\circ C}$ is placed in $0.32\,\mathrm{kg}$ of water at $19\,\mathrm{^\circ C}$. Heat exchange with the cup and surroundings is negligible.
a. Write the energy-balance equation for the final temperature $T_f$.[2]
b. Calculate the equilibrium temperature.[4]
c. State how including the cup would affect the calculated final temperature.[2]
Show complete worked solution
(a)
Energy lost by the hot sample equals energy gained by the water:
$$m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$$
Final answer: $m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$
(b)
$$(0.18)(385)(92-T_f)=(0.32)(4180)(T_f-19)$$
Expanding and collecting $T_f$ terms gives
$$T_f=\boxed{22.6\,\mathrm{^\circ C}}$$
Final answer: $22.6\,\mathrm{^\circ C}$
(c)
The cup also gains energy. The hot sample must warm both the water and the cup, so the equilibrium temperature lies below the value calculated when the cup is ignored.
Final answer: The final temperature would be lower if the cup initially has the water temperature.
QUESTION 63[7 marks]
Multi-partAdvancedCalculatorAbout 8.8 min
A star has luminosity $4.20e+29\,\mathrm{W}$ and is $3.80e+18\,\mathrm{m}$ from Earth.
a. Calculate its apparent brightness at Earth.[3]
b. State why luminosity and apparent brightness are different quantities.[2]
c. Determine the distance at which the apparent brightness would be one quarter as large.[2]
A vacuum flask uses silvered surfaces to control thermal-energy transfer.
a. Identify the principal transfer mechanism affected.[1]
b. Explain the effect using a particle or radiation model.[3]
c. Suggest one measurement that could test the effectiveness of the design.[2]
Show complete worked solution
(a)
The design feature is chosen to reduce radiation. The relevant mechanism is therefore radiation.
Final answer: Radiation.
(b)
Silvered surfaces reduce radiation. The explanation follows from particle collisions and bulk fluid motion for conduction/convection, or absorption and emission properties for radiation.
Final answer: Reduce radiation.
(c)
Use identical systems with the same initial temperature difference. Change only the named feature, record temperature at equal time intervals, and compare the cooling or heating rates.
Final answer: Measure temperature change over equal times with and without the feature.
QUESTION 65[6 marks]
Multi-partAdvancedCalculatorAbout 7.5 min
A saucepan uses a metal base to control thermal-energy transfer.
a. Identify the principal transfer mechanism affected.[1]
b. Explain the effect using a particle or radiation model.[3]
c. Suggest one measurement that could test the effectiveness of the design.[2]
Show complete worked solution
(a)
The design feature is chosen to increase conduction. The relevant mechanism is therefore conduction.
Final answer: Conduction.
(b)
A metal base increase conduction. The explanation follows from particle collisions and bulk fluid motion for conduction/convection, or absorption and emission properties for radiation.
Final answer: Increase conduction.
(c)
Use identical systems with the same initial temperature difference. Change only the named feature, record temperature at equal time intervals, and compare the cooling or heating rates.
Final answer: Measure temperature change over equal times with and without the feature.
QUESTION 66[7 marks]
Multi-partAdvancedCalculatorAbout 8.8 min
An ideal gas is at $285\,\mathrm{K}$. One molecule has mass $4.650e-26\,\mathrm{kg}$.
a. Calculate the mean translational kinetic energy of one molecule.[3]
b. Estimate the molecular speed corresponding to this kinetic energy.[3]
c. State how the mean kinetic energy changes if the absolute temperature doubles.[1]
$\bar E_k\propto T$, so doubling $T$ doubles $\bar E_k$.
Final answer: It doubles.
QUESTION 67[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A temperature sensor records $72.0\,\mathrm{^\circ C}$. It later records a temperature $36.0\,\mathrm{K}$ higher.
a. Convert the initial temperature to kelvin.[2]
b. Determine the final temperature in degrees Celsius.[2]
c. State why kelvin must be used in molecular-energy equations.[2]
Show complete worked solution
(a)
$$T=72.0+273.15=\boxed{345.15\,\mathrm{K}}$$
Final answer: $345.15\,\mathrm{K}$
(b)
A temperature interval has the same numerical value in kelvin and degrees Celsius.
$$\theta_f=72.0+36.0=\boxed{108.0\,\mathrm{^\circ C}}$$
Final answer: $108.0\,\mathrm{^\circ C}$
(c)
Quantities such as average molecular kinetic energy are proportional to absolute temperature. Celsius has an arbitrary zero, so substituting Celsius would not preserve that proportionality.
Final answer: Kelvin is an absolute thermodynamic scale with zero at minimum thermal energy.
QUESTION 68[8 marks]
Multi-partStandardCalculatorAbout 10 min
A room has an opaque wall area of $16\,\mathrm{m^2}$ and a single-glazed window area of $2.6\,\mathrm{m^2}$. The wall has $k=0.72\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.24\,\mathrm{m}$; the glass has $k=1.0\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.008\,\mathrm{m}$. The temperature difference is $13\,\mathrm{K}$.
a. Determine the power conducted through the opaque wall.[3]
b. Determine the power conducted through the window.[3]
c. Identify which part needs improved insulation and justify quantitatively.[2]
$$\frac{P_g}{P_w}=\frac{4225}{624}=6.77$$
The larger transfer rate identifies the dominant path.
Final answer: The window; it has the larger heat-transfer rate.
QUESTION 69[8 marks]
Multi-partStandardCalculatorAbout 10 min
A $0.27\,\mathrm{kg}$ hot sample with specific heat capacity $840\,\mathrm{J\,kg^{-1}K^{-1}}$ at $83\,\mathrm{^\circ C}$ is placed in $0.44\,\mathrm{kg}$ of water at $22\,\mathrm{^\circ C}$. Heat exchange with the cup and surroundings is negligible.
a. Write the energy-balance equation for the final temperature $T_f$.[2]
b. Calculate the equilibrium temperature.[4]
c. State how including the cup would affect the calculated final temperature.[2]
Show complete worked solution
(a)
Energy lost by the hot sample equals energy gained by the water:
$$m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$$
Final answer: $m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$
(b)
$$(0.27)(840)(83-T_f)=(0.44)(4180)(T_f-22)$$
Expanding and collecting $T_f$ terms gives
$$T_f=\boxed{28.7\,\mathrm{^\circ C}}$$
Final answer: $28.7\,\mathrm{^\circ C}$
(c)
The cup also gains energy. The hot sample must warm both the water and the cup, so the equilibrium temperature lies below the value calculated when the cup is ignored.
Final answer: The final temperature would be lower if the cup initially has the water temperature.
QUESTION 70[8 marks]
Data-basedFoundationCalculatorAbout 10 min
The graph shows an idealized heating curve for a $0.29\,\mathrm{kg}$ substance heated at $480\,\mathrm{W}$. In its first phase, $c=2350\,\mathrm{J\,kg^{-1}K^{-1}}$; the first temperature rise is $18\,\mathrm{K}$ and the latent heat for the first plateau is $2.00e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the time for the first temperature rise.[3]
b. Calculate the duration of the first plateau.[3]
c. Explain the sloping and horizontal sections of the graph.[2]
On a slope, energy increases mean molecular kinetic energy, so temperature rises. On a plateau, energy changes molecular separation and potential energy, so temperature stays constant.
Final answer: Slopes show rising mean kinetic energy; plateaus show increasing potential energy during a phase change.
QUESTION 71[7 marks]
Multi-partFoundationCalculatorAbout 8.8 min
An electrical heater of power $62\,\mathrm{W}$ heats a $1.35\,\mathrm{kg}$ water block for $360\,\mathrm{s}$. Use $c=4180\,\mathrm{J\,kg^{-1}K^{-1}}$.
a. Calculate the electrical energy supplied.[2]
b. Predict the temperature rise if no energy is lost.[3]
c. The measured rise is $3.60\,\mathrm{K}$. Calculate the percentage of input energy transferred to the block.[2]
For fixed $m$ and $c$, useful energy is proportional to $\Delta T$.
$$\eta=\frac{3.60}{3.96}\times100=\boxed{91.0\%}$$
Final answer: $91.0\%$
QUESTION 72[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A heater supplies constant power $1100\,\mathrm{W}$ during the vaporization of a refrigerant. The mass changing phase is $0.072\,\mathrm{kg}$ and the specific latent heat is $8.600e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the energy required for the phase change.[2]
b. Determine the minimum time required.[2]
c. Explain why the temperature remains approximately constant during the phase change.[2]
The supplied energy separates or rearranges particles against intermolecular forces. Mean kinetic energy, and therefore temperature, remains approximately constant until the phase change is complete.
Final answer: Energy increases intermolecular potential energy rather than mean molecular kinetic energy.
QUESTION 73[6 marks]
Multi-partStandardCalculatorAbout 7.5 min
A sample of glass has mass $0.580\,\mathrm{kg}$ and volume $232.0\,\mathrm{cm^3}$.
a. Convert the volume to $\mathrm{m^3}$.[2]
b. Calculate the density.[2]
c. Explain, using density, whether it floats in water.[2]
$\rho_{glass}=2500\,\mathrm{kg\,m^{-3}}$ and $\rho_{water}\approx1000\,\mathrm{kg\,m^{-3}}$.
The sample is denser, so its weight exceeds the maximum buoyant force before full support and it sinks.
Final answer: It sinks because its density exceeds that of water.
QUESTION 74[6 marks]
Multi-partFoundationCalculatorAbout 7.5 min
A flat insulating panel has area $0.82\,\mathrm{m^2}$, thickness $0.021\,\mathrm{m}$ and thermal conductivity $0.22\,\mathrm{W\,m^{-1}\,K^{-1}}$. Its faces differ in temperature by $25\,\mathrm{K}$.
a. Calculate the steady rate of thermal-energy transfer.[3]
b. Calculate the energy transferred in $12.0$ minutes.[2]
c. State the effect of doubling the thickness while other quantities remain fixed.[1]
A temperature sensor records $-38.0\,\mathrm{^\circ C}$. It later records a temperature $28.0\,\mathrm{K}$ higher.
a. Convert the initial temperature to kelvin.[2]
b. Determine the final temperature in degrees Celsius.[2]
c. State why kelvin must be used in molecular-energy equations.[2]
Show complete worked solution
(a)
$$T=-38.0+273.15=\boxed{235.15\,\mathrm{K}}$$
Final answer: $235.15\,\mathrm{K}$
(b)
A temperature interval has the same numerical value in kelvin and degrees Celsius.
$$\theta_f=-38.0+28.0=\boxed{-10.0\,\mathrm{^\circ C}}$$
Final answer: $-10.0\,\mathrm{^\circ C}$
(c)
Quantities such as average molecular kinetic energy are proportional to absolute temperature. Celsius has an arbitrary zero, so substituting Celsius would not preserve that proportionality.
Final answer: Kelvin is an absolute thermodynamic scale with zero at minimum thermal energy.
QUESTION 77[6 marks]
Multi-partStandardCalculatorAbout 7.5 min
A sample of water has mass $0.800\,\mathrm{kg}$ and volume $800.0\,\mathrm{cm^3}$.
a. Convert the volume to $\mathrm{m^3}$.[2]
b. Calculate the density.[2]
c. Explain, using density, whether it floats in water.[2]
$\rho_{water}=1000\,\mathrm{kg\,m^{-3}}$ and $\rho_{water}\approx1000\,\mathrm{kg\,m^{-3}}$.
The densities are approximately equal, so it is neutrally buoyant.
Final answer: It is neutrally buoyant in water.
QUESTION 78[6 marks]
Multi-partStandardCalculatorAbout 7.5 min
A sample of copper has mass $0.470\,\mathrm{kg}$ and volume $52.5\,\mathrm{cm^3}$.
a. Convert the volume to $\mathrm{m^3}$.[2]
b. Calculate the density.[2]
c. Explain, using density, whether it floats in water.[2]
$\rho_{copper}=8960\,\mathrm{kg\,m^{-3}}$ and $\rho_{water}\approx1000\,\mathrm{kg\,m^{-3}}$.
The sample is denser, so its weight exceeds the maximum buoyant force before full support and it sinks.
Final answer: It sinks because its density exceeds that of water.
QUESTION 79[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A temperature sensor records $145.0\,\mathrm{^\circ C}$. It later records a temperature $40.0\,\mathrm{K}$ higher.
a. Convert the initial temperature to kelvin.[2]
b. Determine the final temperature in degrees Celsius.[2]
c. State why kelvin must be used in molecular-energy equations.[2]
Show complete worked solution
(a)
$$T=145.0+273.15=\boxed{418.15\,\mathrm{K}}$$
Final answer: $418.15\,\mathrm{K}$
(b)
A temperature interval has the same numerical value in kelvin and degrees Celsius.
$$\theta_f=145.0+40.0=\boxed{185.0\,\mathrm{^\circ C}}$$
Final answer: $185.0\,\mathrm{^\circ C}$
(c)
Quantities such as average molecular kinetic energy are proportional to absolute temperature. Celsius has an arbitrary zero, so substituting Celsius would not preserve that proportionality.
Final answer: Kelvin is an absolute thermodynamic scale with zero at minimum thermal energy.
QUESTION 80[7 marks]
Multi-partFoundationCalculatorAbout 8.8 min
An electrical heater of power $56\,\mathrm{W}$ heats a $1.20\,\mathrm{kg}$ granite block for $330\,\mathrm{s}$. Use $c=790\,\mathrm{J\,kg^{-1}K^{-1}}$.
a. Calculate the electrical energy supplied.[2]
b. Predict the temperature rise if no energy is lost.[3]
c. The measured rise is $17.74\,\mathrm{K}$. Calculate the percentage of input energy transferred to the block.[2]
For fixed $m$ and $c$, useful energy is proportional to $\Delta T$.
$$\eta=\frac{15.48}{17.0}\times100=\boxed{91.0\%}$$
Final answer: $91.0\%$
QUESTION 82[6 marks]
Multi-partFoundationCalculatorAbout 7.5 min
A flat insulating panel has area $0.70\,\mathrm{m^2}$, thickness $0.018\,\mathrm{m}$ and thermal conductivity $0.14\,\mathrm{W\,m^{-1}\,K^{-1}}$. Its faces differ in temperature by $22\,\mathrm{K}$.
a. Calculate the steady rate of thermal-energy transfer.[3]
b. Calculate the energy transferred in $12.0$ minutes.[2]
c. State the effect of doubling the thickness while other quantities remain fixed.[1]
A heater supplies constant power $900\,\mathrm{W}$ during the melting of ice. The mass changing phase is $0.056\,\mathrm{kg}$ and the specific latent heat is $3.340e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the energy required for the phase change.[2]
b. Determine the minimum time required.[2]
c. Explain why the temperature remains approximately constant during the phase change.[2]
The supplied energy separates or rearranges particles against intermolecular forces. Mean kinetic energy, and therefore temperature, remains approximately constant until the phase change is complete.
Final answer: Energy increases intermolecular potential energy rather than mean molecular kinetic energy.
QUESTION 84[7 marks]
Multi-partAdvancedCalculatorAbout 8.8 min
An ideal gas is at $357\,\mathrm{K}$. One molecule has mass $5.850e-26\,\mathrm{kg}$.
a. Calculate the mean translational kinetic energy of one molecule.[3]
b. Estimate the molecular speed corresponding to this kinetic energy.[3]
c. State how the mean kinetic energy changes if the absolute temperature doubles.[1]
$\bar E_k\propto T$, so doubling $T$ doubles $\bar E_k$.
Final answer: It doubles.
QUESTION 85[7 marks]
Multi-partFoundationCalculatorAbout 8.8 min
An electrical heater of power $38\,\mathrm{W}$ heats a $0.75\,\mathrm{kg}$ aluminium block for $240\,\mathrm{s}$. Use $c=900\,\mathrm{J\,kg^{-1}K^{-1}}$.
a. Calculate the electrical energy supplied.[2]
b. Predict the temperature rise if no energy is lost.[3]
c. The measured rise is $12.30\,\mathrm{K}$. Calculate the percentage of input energy transferred to the block.[2]
For fixed $m$ and $c$, useful energy is proportional to $\Delta T$.
$$\eta=\frac{12.30}{13.5}\times100=\boxed{91.0\%}$$
Final answer: $91.0\%$
QUESTION 86[8 marks]
Multi-partStandardCalculatorAbout 10 min
A $0.30\,\mathrm{kg}$ hot sample with specific heat capacity $130\,\mathrm{J\,kg^{-1}K^{-1}}$ at $80\,\mathrm{^\circ C}$ is placed in $0.48\,\mathrm{kg}$ of water at $23\,\mathrm{^\circ C}$. Heat exchange with the cup and surroundings is negligible.
a. Write the energy-balance equation for the final temperature $T_f$.[2]
b. Calculate the equilibrium temperature.[4]
c. State how including the cup would affect the calculated final temperature.[2]
Show complete worked solution
(a)
Energy lost by the hot sample equals energy gained by the water:
$$m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$$
Final answer: $m_hc_h(T_h-T_f)=m_wc_w(T_f-T_w)$
(b)
$$(0.30)(130)(80-T_f)=(0.48)(4180)(T_f-23)$$
Expanding and collecting $T_f$ terms gives
$$T_f=\boxed{24.1\,\mathrm{^\circ C}}$$
Final answer: $24.1\,\mathrm{^\circ C}$
(c)
The cup also gains energy. The hot sample must warm both the water and the cup, so the equilibrium temperature lies below the value calculated when the cup is ignored.
Final answer: The final temperature would be lower if the cup initially has the water temperature.
QUESTION 87[7 marks]
Multi-partFoundationCalculatorAbout 8.8 min
An electrical heater of power $44\,\mathrm{W}$ heats a $0.90\,\mathrm{kg}$ copper block for $270\,\mathrm{s}$. Use $c=385\,\mathrm{J\,kg^{-1}K^{-1}}$.
a. Calculate the electrical energy supplied.[2]
b. Predict the temperature rise if no energy is lost.[3]
c. The measured rise is $31.20\,\mathrm{K}$. Calculate the percentage of input energy transferred to the block.[2]
A heater supplies constant power $800\,\mathrm{W}$ during the vaporization of ethanol. The mass changing phase is $0.048\,\mathrm{kg}$ and the specific latent heat is $8.400e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the energy required for the phase change.[2]
b. Determine the minimum time required.[2]
c. Explain why the temperature remains approximately constant during the phase change.[2]
The supplied energy separates or rearranges particles against intermolecular forces. Mean kinetic energy, and therefore temperature, remains approximately constant until the phase change is complete.
Final answer: Energy increases intermolecular potential energy rather than mean molecular kinetic energy.
QUESTION 90[6 marks]
Multi-partStandardCalculatorAbout 7.5 min
A sample of granite has mass $0.690\,\mathrm{kg}$ and volume $250.9\,\mathrm{cm^3}$.
a. Convert the volume to $\mathrm{m^3}$.[2]
b. Calculate the density.[2]
c. Explain, using density, whether it floats in water.[2]
$\rho_{granite}=2750\,\mathrm{kg\,m^{-3}}$ and $\rho_{water}\approx1000\,\mathrm{kg\,m^{-3}}$.
The sample is denser, so its weight exceeds the maximum buoyant force before full support and it sinks.
Final answer: It sinks because its density exceeds that of water.
QUESTION 91[6 marks]
Multi-partFoundationCalculatorAbout 7.5 min
A flat insulating panel has area $1.18\,\mathrm{m^2}$, thickness $0.030\,\mathrm{m}$ and thermal conductivity $0.60\,\mathrm{W\,m^{-1}\,K^{-1}}$. Its faces differ in temperature by $34\,\mathrm{K}$.
a. Calculate the steady rate of thermal-energy transfer.[3]
b. Calculate the energy transferred in $12.0$ minutes.[2]
c. State the effect of doubling the thickness while other quantities remain fixed.[1]
A flat insulating panel has area $1.06\,\mathrm{m^2}$, thickness $0.027\,\mathrm{m}$ and thermal conductivity $1.40\,\mathrm{W\,m^{-1}\,K^{-1}}$. Its faces differ in temperature by $31\,\mathrm{K}$.
a. Calculate the steady rate of thermal-energy transfer.[3]
b. Calculate the energy transferred in $12.0$ minutes.[2]
c. State the effect of doubling the thickness while other quantities remain fixed.[1]
A cylindrical heating element of radius $0.80\,\mathrm{mm}$ and length $0.73\,\mathrm{m}$ is at $745\,\mathrm{K}$. Its emissivity is $0.82$; neglect radiation from its ends.
a. Calculate its radiating surface area.[2]
b. Calculate the emitted radiation power.[3]
c. Determine the factor change in power if the absolute temperature rises by $10\%$.[2]
The graph shows an idealized heating curve for a $0.41\,\mathrm{kg}$ substance heated at $660\,\mathrm{W}$. In its first phase, $c=3100\,\mathrm{J\,kg^{-1}K^{-1}}$; the first temperature rise is $18\,\mathrm{K}$ and the latent heat for the first plateau is $2.60e+05\,\mathrm{J\,kg^{-1}}$.
a. Calculate the time for the first temperature rise.[3]
b. Calculate the duration of the first plateau.[3]
c. Explain the sloping and horizontal sections of the graph.[2]
On a slope, energy increases mean molecular kinetic energy, so temperature rises. On a plateau, energy changes molecular separation and potential energy, so temperature stays constant.
Final answer: Slopes show rising mean kinetic energy; plateaus show increasing potential energy during a phase change.
QUESTION 95[8 marks]
Multi-partStandardCalculatorAbout 10 min
A room has an opaque wall area of $18\,\mathrm{m^2}$ and a single-glazed window area of $3.0\,\mathrm{m^2}$. The wall has $k=0.72\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.26\,\mathrm{m}$; the glass has $k=1.0\,\mathrm{W\,m^{-1}K^{-1}}$ and thickness $0.010\,\mathrm{m}$. The temperature difference is $17\,\mathrm{K}$.
a. Determine the power conducted through the opaque wall.[3]
b. Determine the power conducted through the window.[3]
c. Identify which part needs improved insulation and justify quantitatively.[2]
$$\frac{P_g}{P_w}=\frac{5100}{847}=6.02$$
The larger transfer rate identifies the dominant path.
Final answer: The window; it has the larger heat-transfer rate.
QUESTION 96[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A heater supplies constant power $700\,\mathrm{W}$ during the vaporization of water. The mass changing phase is $0.040\,\mathrm{kg}$ and the specific latent heat is $2.260e+06\,\mathrm{J\,kg^{-1}}$.
a. Calculate the energy required for the phase change.[2]
b. Determine the minimum time required.[2]
c. Explain why the temperature remains approximately constant during the phase change.[2]
The supplied energy separates or rearranges particles against intermolecular forces. Mean kinetic energy, and therefore temperature, remains approximately constant until the phase change is complete.
Final answer: Energy increases intermolecular potential energy rather than mean molecular kinetic energy.
QUESTION 97[6 marks]
Multi-partAppliedCalculatorAbout 7.5 min
A temperature sensor records $660.0\,\mathrm{^\circ C}$. It later records a temperature $44.0\,\mathrm{K}$ higher.
a. Convert the initial temperature to kelvin.[2]
b. Determine the final temperature in degrees Celsius.[2]
c. State why kelvin must be used in molecular-energy equations.[2]
Show complete worked solution
(a)
$$T=660.0+273.15=\boxed{933.15\,\mathrm{K}}$$
Final answer: $933.15\,\mathrm{K}$
(b)
A temperature interval has the same numerical value in kelvin and degrees Celsius.
$$\theta_f=660.0+44.0=\boxed{704.0\,\mathrm{^\circ C}}$$
Final answer: $704.0\,\mathrm{^\circ C}$
(c)
Quantities such as average molecular kinetic energy are proportional to absolute temperature. Celsius has an arbitrary zero, so substituting Celsius would not preserve that proportionality.
Final answer: Kelvin is an absolute thermodynamic scale with zero at minimum thermal energy.
QUESTION 98[7 marks]
Multi-partAppliedCalculatorAbout 8.8 min
A cylindrical heating element of radius $0.70\,\mathrm{mm}$ and length $0.65\,\mathrm{m}$ is at $700\,\mathrm{K}$. Its emissivity is $0.78$; neglect radiation from its ends.
a. Calculate its radiating surface area.[2]
b. Calculate the emitted radiation power.[3]
c. Determine the factor change in power if the absolute temperature rises by $10\%$.[2]
$\bar E_k\propto T$, so doubling $T$ doubles $\bar E_k$.
Final answer: It doubles.
QUESTION 100[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
An ideal gas is at $285\,\mathrm{K}$. One molecule has mass $4.650e-26\,\mathrm{kg}$.
Calculate the mean translational kinetic energy of one molecule.
Show complete worked solution
Correct answer: A
$$\bar E_k=\frac32k_BT=\frac32(1.38\times10^{-23})(285)=\boxed{5.90\times10^{-21}\,\mathrm{J}}$$
Therefore the correct option is A.
QUESTION 101[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A star has luminosity $2.80e+29\,\mathrm{W}$ and is $3.00e+18\,\mathrm{m}$ from Earth.
Calculate its apparent brightness at Earth.
Show complete worked solution
Correct answer: B
$$b=\frac{L}{4\pi d^2}=\frac{2.80e+29}{4\pi(3.00e+18)^2}=\boxed{2.48\times10^{-9}\,\mathrm{W\,m^{-2}}}$$
Therefore the correct option is B.
QUESTION 102[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
An ideal gas is at $303\,\mathrm{K}$. One molecule has mass $4.950e-26\,\mathrm{kg}$.
Calculate the mean translational kinetic energy of one molecule.
Show complete worked solution
Correct answer: C
$$\bar E_k=\frac32k_BT=\frac32(1.38\times10^{-23})(303)=\boxed{6.27\times10^{-21}\,\mathrm{J}}$$
Therefore the correct option is C.
QUESTION 103[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A star has luminosity $3.50e+29\,\mathrm{W}$ and is $3.40e+18\,\mathrm{m}$ from Earth.
Calculate its apparent brightness at Earth.
Show complete worked solution
Correct answer: D
$$b=\frac{L}{4\pi d^2}=\frac{3.50e+29}{4\pi(3.40e+18)^2}=\boxed{2.41\times10^{-9}\,\mathrm{W\,m^{-2}}}$$
Therefore the correct option is D.
QUESTION 104[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
An ideal gas is at $321\,\mathrm{K}$. One molecule has mass $5.250e-26\,\mathrm{kg}$.
Calculate the mean translational kinetic energy of one molecule.
Show complete worked solution
Correct answer: A
$$\bar E_k=\frac32k_BT=\frac32(1.38\times10^{-23})(321)=\boxed{6.64\times10^{-21}\,\mathrm{J}}$$
Therefore the correct option is A.
QUESTION 105[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A star has luminosity $4.20e+29\,\mathrm{W}$ and is $3.80e+18\,\mathrm{m}$ from Earth.
Calculate its apparent brightness at Earth.
Show complete worked solution
Correct answer: B
$$b=\frac{L}{4\pi d^2}=\frac{4.20e+29}{4\pi(3.80e+18)^2}=\boxed{2.31\times10^{-9}\,\mathrm{W\,m^{-2}}}$$
Therefore the correct option is B.
QUESTION 106[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
An ideal gas is at $339\,\mathrm{K}$. One molecule has mass $5.550e-26\,\mathrm{kg}$.
Calculate the mean translational kinetic energy of one molecule.
Show complete worked solution
Correct answer: C
$$\bar E_k=\frac32k_BT=\frac32(1.38\times10^{-23})(339)=\boxed{7.02\times10^{-21}\,\mathrm{J}}$$
Therefore the correct option is C.
QUESTION 107[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A star has luminosity $4.90e+29\,\mathrm{W}$ and is $4.20e+18\,\mathrm{m}$ from Earth.
Calculate its apparent brightness at Earth.
Show complete worked solution
Correct answer: D
$$b=\frac{L}{4\pi d^2}=\frac{4.90e+29}{4\pi(4.20e+18)^2}=\boxed{2.21\times10^{-9}\,\mathrm{W\,m^{-2}}}$$
Therefore the correct option is D.
QUESTION 108[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
An ideal gas is at $357\,\mathrm{K}$. One molecule has mass $5.850e-26\,\mathrm{kg}$.
Calculate the mean translational kinetic energy of one molecule.
Show complete worked solution
Correct answer: A
$$\bar E_k=\frac32k_BT=\frac32(1.38\times10^{-23})(357)=\boxed{7.39\times10^{-21}\,\mathrm{J}}$$
Therefore the correct option is A.
QUESTION 109[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A star has luminosity $5.60e+29\,\mathrm{W}$ and is $4.60e+18\,\mathrm{m}$ from Earth.
Calculate its apparent brightness at Earth.
Show complete worked solution
Correct answer: B
$$b=\frac{L}{4\pi d^2}=\frac{5.60e+29}{4\pi(4.60e+18)^2}=\boxed{2.11\times10^{-9}\,\mathrm{W\,m^{-2}}}$$
Therefore the correct option is B.
QUESTION 110[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A cylindrical heating element of radius $0.70\,\mathrm{mm}$ and length $0.65\,\mathrm{m}$ is at $700\,\mathrm{K}$. Its emissivity is $0.78$; neglect radiation from its ends.
Calculate the emitted radiation power.
Show complete worked solution
Correct answer: C
$$P=e\sigma AT^4=(0.78)(5.67\times10^{-8})(2.86\times10^{-3})(700)^4=\boxed{30.4\,\mathrm{W}}$$
Therefore the correct option is C.
QUESTION 111[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A cylindrical heating element of radius $0.80\,\mathrm{mm}$ and length $0.73\,\mathrm{m}$ is at $745\,\mathrm{K}$. Its emissivity is $0.82$; neglect radiation from its ends.
Calculate the emitted radiation power.
Show complete worked solution
Correct answer: D
$$P=e\sigma AT^4=(0.82)(5.67\times10^{-8})(3.67\times10^{-3})(745)^4=\boxed{52.6\,\mathrm{W}}$$
Therefore the correct option is D.
QUESTION 112[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A cylindrical heating element of radius $0.90\,\mathrm{mm}$ and length $0.81\,\mathrm{m}$ is at $790\,\mathrm{K}$. Its emissivity is $0.86$; neglect radiation from its ends.
Calculate the emitted radiation power.
Show complete worked solution
Correct answer: A
$$P=e\sigma AT^4=(0.86)(5.67\times10^{-8})(4.58\times10^{-3})(790)^4=\boxed{87.0\,\mathrm{W}}$$
Therefore the correct option is A.
QUESTION 113[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A cylindrical heating element of radius $1.00\,\mathrm{mm}$ and length $0.89\,\mathrm{m}$ is at $835\,\mathrm{K}$. Its emissivity is $0.90$; neglect radiation from its ends.
Calculate the emitted radiation power.
Show complete worked solution
Correct answer: B
$$P=e\sigma AT^4=(0.90)(5.67\times10^{-8})(5.59\times10^{-3})(835)^4=\boxed{139\,\mathrm{W}}$$
Therefore the correct option is B.
QUESTION 114[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A cylindrical heating element of radius $1.10\,\mathrm{mm}$ and length $0.97\,\mathrm{m}$ is at $880\,\mathrm{K}$. Its emissivity is $0.94$; neglect radiation from its ends.
Calculate the emitted radiation power.
Show complete worked solution
Correct answer: C
$$P=e\sigma AT^4=(0.94)(5.67\times10^{-8})(6.70\times10^{-3})(880)^4=\boxed{214\,\mathrm{W}}$$
Therefore the correct option is C.
QUESTION 115[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A temperature sensor records $-38.0\,\mathrm{^\circ C}$. It later records a temperature $28.0\,\mathrm{K}$ higher.
Convert the initial temperature to kelvin.
Show complete worked solution
Correct answer: D
$$T=-38.0+273.15=\boxed{235.15\,\mathrm{K}}$$
Therefore the correct option is D.
QUESTION 116[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A heater supplies constant power $700\,\mathrm{W}$ during the vaporization of water. The mass changing phase is $0.040\,\mathrm{kg}$ and the specific latent heat is $2.260e+06\,\mathrm{J\,kg^{-1}}$.
Calculate the energy required for the phase change.
Show complete worked solution
Correct answer: A
$$Q=mL=(0.040)(2.260e+06)=\boxed{9.04\times10^{4}\,\mathrm{J}}$$
Therefore the correct option is A.
QUESTION 117[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A temperature sensor records $18.0\,\mathrm{^\circ C}$. It later records a temperature $32.0\,\mathrm{K}$ higher.
Convert the initial temperature to kelvin.
Show complete worked solution
Correct answer: B
$$T=18.0+273.15=\boxed{291.15\,\mathrm{K}}$$
Therefore the correct option is B.
QUESTION 118[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A heater supplies constant power $800\,\mathrm{W}$ during the vaporization of ethanol. The mass changing phase is $0.048\,\mathrm{kg}$ and the specific latent heat is $8.400e+05\,\mathrm{J\,kg^{-1}}$.
Calculate the energy required for the phase change.
Show complete worked solution
Correct answer: C
$$Q=mL=(0.048)(8.400e+05)=\boxed{4.03\times10^{4}\,\mathrm{J}}$$
Therefore the correct option is C.
QUESTION 119[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A temperature sensor records $72.0\,\mathrm{^\circ C}$. It later records a temperature $36.0\,\mathrm{K}$ higher.
Convert the initial temperature to kelvin.
Show complete worked solution
Correct answer: D
$$T=72.0+273.15=\boxed{345.15\,\mathrm{K}}$$
Therefore the correct option is D.