IB Physics Quiz: Apply Thermal Energy Transfers
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Apply Thermal Energy TransfersQuestion 1 of 20

Two spherical stars, P and Q, radiate as black bodies. The radius of star P is twice the radius of star Q. The absolute surface temperature of star P is half the surface temperature of star Q. What is the ratio of the power radiated by P to the power radiated by Q?

1/4
1/2
1
2
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IB Physics Quiz

IB Physics Quiz: Apply Thermal Energy Transfers

Practice Apply Thermal Energy Transfers in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Thermal Energy Transfers, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

Two spherical stars, P and Q, radiate as black bodies. The radius of star P is twice the radius of star Q. The absolute surface temperature of star P is half the surface temperature of star Q. What is the ratio of the power radiated by P to the power radiated by Q?

  1. 1/4 (correct answer)
  2. 1/2
  3. 1
  4. 2
Explanation: The power radiated by a star (luminosity) is given by the Stefan-Boltzmann law, L=σAT4L = \sigma A T^4, where A is the surface area. For a sphere, A=4πr2A = 4\pi r^2. So, L∝r2T4L \propto r^2 T^4. Let rQr_Q and TQT_Q be the radius and temperature of star Q. Then rP=2rQr_P = 2r_Q and TP=0.5TQT_P = 0.5T_Q. The ratio of the powers is LPLQ=rP2TP4rQ2TQ4=(2rQ)2(0.5TQ)4rQ2TQ4=4rQ2⋅(1/16)TQ4rQ2TQ4=416=14\frac{L_P}{L_Q} = \frac{r_P^2 T_P^4}{r_Q^2 T_Q^4} = \frac{(2r_Q)^2 (0.5T_Q)^4}{r_Q^2 T_Q^4} = \frac{4r_Q^2 \cdot (1/16)T_Q^4}{r_Q^2 T_Q^4} = \frac{4}{16} = \frac{1}{4}.

Question 2

The surface of a hot object radiates thermal energy. Due to a change in its temperature, the peak wavelength of its emitted radiation decreases by 20%. By what factor does the total power radiated per unit area from the object's surface increase?

  1. 1.25
  2. 1.56
  3. 2.44 (correct answer)
  4. 3.05
Explanation: According to Wien's displacement law, λmaxT=constant\lambda_{max} T = \text{constant}, so temperature T is inversely proportional to the peak wavelength λmax\lambda_{max}. If λmax\lambda_{max} decreases by 20%, the new wavelength is λmax′=0.8λmax\lambda'_{max} = 0.8 \lambda_{max}. The new temperature T′T' is therefore T′=T/0.8=1.25TT' = T / 0.8 = 1.25T. According to the Stefan-Boltzmann law, the power radiated per unit area is P/A=σT4P/A = \sigma T^4. The new power per unit area will be proportional to (T′)4(T')^4. The factor of increase is (T′)4T4=(T′T)4=(1.25)4≈2.44\frac{(T')^4}{T^4} = (\frac{T'}{T})^4 = (1.25)^4 \approx 2.44.

Question 3

A heater with a constant power output is used to first melt a 1 kg block of a substance at its melting point, which takes a time tmt_m. The same heater is then used to raise the temperature of the resulting liquid by 50 K, which takes a time tht_h. The ratio tm/tht_m / t_h is found to be 4.0. What is the specific heat capacity of the substance in its liquid phase? (Specific latent heat of fusion = 3.0×1053.0 \times 10^5 J kg⁻¹)

  1. 1.5 × 10³ J kg⁻¹ K⁻¹ (correct answer)
  2. 2.5 × 10³ J kg⁻¹ K⁻¹
  3. 4.0 × 10³ J kg⁻¹ K⁻¹
  4. 6.0 × 10³ J kg⁻¹ K⁻¹
Explanation: Let P be the constant power of the heater. The energy required to melt the substance is Qm=mLf=PtmQ_m = mL_f = P t_m. The energy required to heat the liquid is Qh=mcΔT=PthQ_h = mc\Delta T = P t_h. Taking the ratio of these two equations: QmQh=PtmPth=tmth\frac{Q_m}{Q_h} = \frac{P t_m}{P t_h} = \frac{t_m}{t_h}. So, mLfmcΔT=tmth\frac{mL_f}{mc\Delta T} = \frac{t_m}{t_h}. The mass m cancels. LfcΔT=tmth\frac{L_f}{c\Delta T} = \frac{t_m}{t_h}. We are given Lf=3.0×105L_f = 3.0 \times 10^5 J kg⁻¹, ΔT=50\Delta T = 50 K, and tm/th=4.0t_m / t_h = 4.0. Substituting these values: 3.0×105c⋅50=4.0\frac{3.0 \times 10^5}{c \cdot 50} = 4.0. Rearranging for c: c=3.0×10550⋅4.0=3.0×105200=1500=1.5×103c = \frac{3.0 \times 10^5}{50 \cdot 4.0} = \frac{3.0 \times 10^5}{200} = 1500 = 1.5 \times 10^3 J kg⁻¹ K⁻¹.

Question 4

On a cold day, a person touches a piece of metal and a piece of wood that have both been outside for a long time. The metal feels significantly colder than the wood. What is the reason for this observation?

  1. The metal has a lower temperature than the wood because it cools down faster.
  2. The metal has a higher thermal conductivity, causing a greater rate of heat transfer from the hand. (correct answer)
  3. The metal has a lower specific heat capacity, so it absorbs heat from the hand more readily.
  4. Convection currents are established more easily in the metal than in the wood.
Explanation: Since both objects have been outside for a long time, they are in thermal equilibrium with the surroundings and thus have the same temperature. The sensation of 'cold' is related to the rate at which heat is transferred away from the skin. Metal has a much higher thermal conductivity than wood. Therefore, it conducts heat away from the hand at a much higher rate, leading to the sensation of it being colder.

Question 5

A vacuum flask (thermos) is designed to minimize heat transfer. It consists of a double-walled glass container with a vacuum between the walls, and the surfaces facing the vacuum are silvered.

  1. The vacuum prevents conduction and convection; the silvering minimizes heat transfer by radiation. (correct answer)
  2. The vacuum prevents radiation; the silvering minimizes heat transfer by conduction and convection.
  3. The vacuum prevents conduction and radiation; the silvering minimizes heat transfer by convection.
  4. The vacuum prevents convection; the silvering minimizes heat transfer by conduction and radiation.
Explanation: Heat can be transferred by conduction, convection, and radiation. A vacuum is a space devoid of particles, so it cannot support heat transfer by conduction (which requires a medium) or convection (which requires fluid movement). The silvered surfaces have low emissivity and are highly reflective. This minimizes heat transfer by radiation, both from entering the flask and from leaving it. Therefore, the vacuum addresses conduction and convection, while the silvering addresses radiation.

Question 6

A glass window has an area A, a thickness x, and a thermal conductivity k. The temperature difference between the inside and outside is ΔT\Delta T, resulting in a rate of heat loss P. What will be the new rate of heat loss if the area is halved, the thickness is doubled, and the temperature difference is doubled?

  1. P/4
  2. P/2 (correct answer)
  3. P
  4. 2P
Explanation: The rate of heat loss by conduction is given by the formula P=kAΔTxP = \frac{kA\Delta T}{x}. Let the new parameters be A′=A/2A' = A/2, x′=2xx' = 2x, and ΔT′=2ΔT\Delta T' = 2\Delta T. The new rate of heat loss, P′P', will be P′=kA′ΔT′x′=k(A/2)(2ΔT)2x=kAΔT2x=12(kAΔTx)=P2P' = \frac{k A' \Delta T'}{x'} = \frac{k (A/2) (2\Delta T)}{2x} = \frac{k A \Delta T}{2x} = \frac{1}{2} \left( \frac{kA\Delta T}{x} \right) = \frac{P}{2}.

Question 7

An insulated container is divided into two compartments by a thermally conducting partition. One compartment contains an ideal monatomic gas X, and the other contains an ideal monatomic gas Y. The mass of a molecule of Y is four times the mass of a molecule of X (mY=4mXm_Y = 4m_X). The system is allowed to reach thermal equilibrium. What is the ratio of the root-mean-square (rms) speed of molecules in X to the rms speed of molecules in Y, vrms,X/vrms,Yv_{rms,X} / v_{rms,Y}?

  1. 1/4
  2. 1/2
  3. 2 (correct answer)
  4. 4
Explanation: At thermal equilibrium, both gases have the same temperature. The average translational kinetic energy of the molecules is the same for both gases: Ek,X=Ek,YE_{k,X} = E_{k,Y}. The average kinetic energy is related to the rms speed by Ek=12mvrms2E_k = \frac{1}{2} m v_{rms}^2. Therefore, 12mXvrms,X2=12mYvrms,Y2\frac{1}{2} m_X v_{rms,X}^2 = \frac{1}{2} m_Y v_{rms,Y}^2. Substituting mY=4mXm_Y = 4m_X gives mXvrms,X2=4mXvrms,Y2m_X v_{rms,X}^2 = 4m_X v_{rms,Y}^2. The mass mXm_X cancels out, leaving vrms,X2=4vrms,Y2v_{rms,X}^2 = 4v_{rms,Y}^2. Taking the square root of both sides gives vrms,X=2vrms,Yv_{rms,X} = 2v_{rms,Y}. The ratio is vrms,X/vrms,Y=2v_{rms,X} / v_{rms,Y} = 2.

Question 8

A thick layer of ice has formed on a lake where the air temperature is constantly below 0°C. The water below the ice is at 0°C. Assuming a steady state, how does the rate at which the thickness of the ice increases, dxdt\frac{dx}{dt}, depend on the current thickness of the ice, xx?

  1. The rate is constant and independent of x.
  2. The rate is directly proportional to x.
  3. The rate is directly proportional to x².
  4. The rate is inversely proportional to x. (correct answer)
Explanation: For the ice to get thicker, heat must be removed from the water at the ice-water interface and conducted through the existing ice layer to the colder air above. The rate of heat conduction is given by P=ΔQΔt=kAΔTxP = \frac{\Delta Q}{\Delta t} = \frac{kA\Delta T}{x}, where x is the thickness of the ice. The heat ΔQ\Delta Q that must be removed to freeze an additional thin layer of thickness dxdx is ΔQ=LfρAdx\Delta Q = L_f \rho A dx, where LfL_f is the latent heat of fusion and ρ\rho is the density of ice. Therefore, the rate of heat removal is ΔQΔt=LfρAdxdt\frac{\Delta Q}{\Delta t} = L_f \rho A \frac{dx}{dt}. Equating the two expressions for the rate of heat flow: LfρAdxdt=kAΔTxL_f \rho A \frac{dx}{dt} = \frac{kA\Delta T}{x}. Rearranging for the rate of thickness increase gives dxdt=(kΔTLfρ)1x\frac{dx}{dt} = \left( \frac{k\Delta T}{L_f \rho} \right) \frac{1}{x}. Since all terms in the parenthesis are constant, the rate dxdt\frac{dx}{dt} is inversely proportional to the thickness x.

Question 9

A 500 g block of an unknown metal at 250°C is placed in a well-insulated container holding 500 g of water at 20.0°C. The system reaches a final equilibrium temperature of 40.0°C. What is the specific heat capacity of the metal? (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹)

  1. 200 J kg⁻¹ K⁻¹
  2. 400 J kg⁻¹ K⁻¹ (correct answer)
  3. 800 J kg⁻¹ K⁻¹
  4. 1600 J kg⁻¹ K⁻¹
Explanation: The heat lost by the metal must equal the heat gained by the water. Let cmc_m be the specific heat capacity of the metal. Qlost=mmcmΔTmQ_{lost} = m_m c_m \Delta T_m and Qgained=mwcwΔTwQ_{gained} = m_w c_w \Delta T_w. The masses are equal (mm=mw=0.5m_m = m_w = 0.5 kg), so they will cancel. cm(Tinitial,m−Tf)=cw(Tf−Tinitial,w)c_m (T_{initial,m} - T_f) = c_w (T_f - T_{initial,w}). cm(250−40)=4200(40.0−20.0)c_m (250 - 40) = 4200 (40.0 - 20.0). cm(210)=4200(20)c_m (210) = 4200 (20). cm=4200×20210=84000210=400c_m = \frac{4200 \times 20}{210} = \frac{84000}{210} = 400 J kg⁻¹ K⁻¹.

Question 10

A blacksmith heats a piece of iron until it glows. Initially, it glows dull red. As it gets hotter, it glows bright orange, and then appears white-hot. Which physics principle best explains this change in the observed colour?

  1. Stefan-Boltzmann law, because the total power radiated increases with temperature.
  2. Wien's displacement law, because the peak wavelength of emission shifts to shorter wavelengths. (correct answer)
  3. Specific heat capacity, because different amounts of energy are needed to change its colour.
  4. Thermal conductivity, because heat spreads through the iron causing it to glow uniformly.
Explanation: The colour of a glowing hot object is determined by the distribution of wavelengths in its emitted thermal radiation. Wien's displacement law (λmaxT=constant\lambda_{max} T = \text{constant}) states that as the temperature (T) of an object increases, the peak wavelength (λmax\lambda_{max}) of its emitted radiation shifts to shorter wavelengths. The peak moves from the infrared, into the red part of the visible spectrum, then towards orange, yellow, and eventually blue. When the object emits strongly across the entire visible spectrum, the combination of colours appears white to the human eye. The Stefan-Boltzmann law describes the total power (brightness), not the colour.

Question 11

The apparent brightness of Star X as measured from Earth is equal to that of Star Y. However, Star X is known to be 9 times farther away than Star Y. What is the ratio of the luminosity (total power output) of Star X to that of Star Y, LX/LYL_X / L_Y?

  1. 1/81
  2. 1/9
  3. 9
  4. 81 (correct answer)
Explanation: Apparent brightness (b) is related to luminosity (L) and distance (d) by the formula b=L4πd2b = \frac{L}{4\pi d^2}. We are given that bX=bYb_X = b_Y and dX=9dYd_X = 9d_Y. Therefore, LX4πdX2=LY4πdY2\frac{L_X}{4\pi d_X^2} = \frac{L_Y}{4\pi d_Y^2}. Rearranging for the ratio of luminosities: LXLY=dX2dY2=(9dY)2dY2=81dY2dY2=81\frac{L_X}{L_Y} = \frac{d_X^2}{d_Y^2} = \frac{(9d_Y)^2}{d_Y^2} = \frac{81d_Y^2}{d_Y^2} = 81.

Question 12

A composite rod is made of two sections of equal length and equal cross-sectional area, joined end-to-end. The thermal conductivity of material A is twice that of material B (kA=2kBk_A = 2k_B). The free end of A is held at 100°C and the free end of B is held at 0°C. What is the temperature at the junction between A and B when the rod is in a steady state of heat flow?

  1. 33°C
  2. 50°C
  3. 67°C (correct answer)
  4. 75°C
Explanation: In a steady state, the rate of heat flow (P=ΔQΔtP = \frac{\Delta Q}{\Delta t}) must be the same through both sections. The formula for heat flow is P=kAΔTxP = kA\frac{\Delta T}{x}. Let TjT_j be the junction temperature. For section A: PA=kAA100−TjLP_A = k_A A \frac{100 - T_j}{L}. For section B: PB=kBATj−0LP_B = k_B A \frac{T_j - 0}{L}. Since PA=PBP_A = P_B and areas (A) and lengths (L) are equal: kA(100−Tj)=kBTjk_A (100 - T_j) = k_B T_j. Substituting kA=2kBk_A = 2k_B: 2kB(100−Tj)=kBTj2k_B (100 - T_j) = k_B T_j. The term kBk_B cancels out. 2(100−Tj)=Tj2(100 - T_j) = T_j. 200−2Tj=Tj200 - 2T_j = T_j. 200=3Tj200 = 3T_j. Tj=200/3≈67T_j = 200/3 \approx 67°C.

Question 13

A sealed container holds a mixture of monatomic argon gas (Ar) and diatomic nitrogen gas (N₂) in thermal equilibrium. The mass of an argon atom is greater than the mass of a nitrogen molecule. Which of the following comparisons is correct?

  1. The average kinetic energy of Ar atoms is greater than that of N₂ molecules.
  2. The average speed of Ar atoms is the same as that of N₂ molecules.
  3. The average kinetic energy of Ar atoms is the same as that of N₂ molecules. (correct answer)
  4. The internal energy of the argon gas is equal to the internal energy of the nitrogen gas.
Explanation: When two gases are in thermal equilibrium, they are at the same temperature (T). The average translational kinetic energy of a molecule is given by Ek=32kBTE_k = \frac{3}{2} k_B T, which depends only on temperature. Therefore, the average kinetic energy of the argon atoms and nitrogen molecules must be the same. Because Ek=12mv2E_k = \frac{1}{2}mv^2 and the masses are different, their average speeds must be different. Internal energy depends on the number of particles and degrees of freedom, which are not given, so we cannot conclude they are equal.

Question 14

A solid substance is heated by a source of constant power. It takes 3.0 minutes to melt the substance completely at its melting point. It then takes 2.0 minutes to raise the temperature of the liquid from the melting point to the boiling point, a change of 100 K. What is the ratio of the specific latent heat of fusion (LfL_f) to the specific heat capacity of the liquid (clc_l)?

  1. 67 K
  2. 100 K
  3. 150 K (correct answer)
  4. 300 K
Explanation: Let P be the power of the source. The energy supplied for melting is Qm=mLf=P×tmQ_m = mL_f = P \times t_m, where tm=3.0t_m = 3.0 min. The energy supplied for heating the liquid is Qh=mclΔT=P×thQ_h = mc_l\Delta T = P \times t_h, where th=2.0t_h = 2.0 min and ΔT=100\Delta T = 100 K. We can write the ratio mLfmclΔT=PtmPth\frac{mL_f}{mc_l\Delta T} = \frac{P t_m}{P t_h}. The mass m and power P cancel out. LfclΔT=tmth\frac{L_f}{c_l\Delta T} = \frac{t_m}{t_h}. We need to find the ratio Lfcl\frac{L_f}{c_l}. Rearranging gives Lfcl=ΔT×tmth=100 K×3.0 min2.0 min=100×1.5=150\frac{L_f}{c_l} = \Delta T \times \frac{t_m}{t_h} = 100 \text{ K} \times \frac{3.0 \text{ min}}{2.0 \text{ min}} = 100 \times 1.5 = 150 K.

Question 15

Two solid objects, X and Y, of equal mass are supplied with thermal energy at the same constant rate. The temperature of X increases at a rate of 4.0 K s⁻¹. The temperature of Y increases at a rate of 2.0 K s⁻¹. Both objects remain in the solid phase. What is the ratio of the specific heat capacity of X to that of Y, cX/cYc_X / c_Y?

  1. 0.25
  2. 0.50 (correct answer)
  3. 2.0
  4. 4.0
Explanation: The rate of energy supply, Power (P), is given by P=dQdtP = \frac{dQ}{dt}. Since Q=mcΔTQ = mc\Delta T, we can write P=mcdTdtP = mc\frac{d T}{dt}. Both objects have the same mass (m) and are supplied with energy at the same rate (P). Therefore, mcX(dTdt)X=mcY(dTdt)Ym c_X (\frac{dT}{dt})_X = m c_Y (\frac{dT}{dt})_Y. The mass m cancels out. cX(4.0 K s⁻¹)=cY(2.0 K s⁻¹)c_X (4.0 \text{ K s⁻¹}) = c_Y (2.0 \text{ K s⁻¹}). Rearranging for the ratio cX/cYc_X / c_Y gives cXcY=2.04.0=0.50\frac{c_X}{c_Y} = \frac{2.0}{4.0} = 0.50.

Question 16

A metal sphere is heated uniformly from 300 K to 600 K. During this process, its volume increases by 1.8%. If the sphere is then cooled back to 300 K while maintaining constant pressure, what is the coefficient of linear expansion of the metal?

  1. 2.0 × 10⁻⁵ K⁻¹ (correct answer)
  2. 6.0 × 10⁻⁵ K⁻¹
  3. 1.8 × 10⁻⁵ K⁻¹
  4. 5.4 × 10⁻⁵ K⁻¹
Explanation: Volume expansion: ΔVV0=βΔT=3αΔT\frac{\Delta V}{V_0} = \beta \Delta T = 3\alpha \Delta T where α\alpha is linear expansion coefficient. Given: ΔVV0=0.018\frac{\Delta V}{V_0} = 0.018 and ΔT=300K\Delta T = 300 K. Therefore: 0.018=3α×3000.018 = 3\alpha \times 300, so α=0.018900=2.0×10−5K−1\alpha = \frac{0.018}{900} = 2.0 \times 10^{-5} K^{-1}. Choice B uses β\beta instead of α\alpha, C confuses percentage with coefficient, D uses incorrect relationship between volume and linear expansion.

Question 17

A blackbody radiator has a surface temperature of 727°C. If its temperature is increased to 1227°C while maintaining the same surface area, by what factor does its total radiated power increase?

  1. 2.0
  2. 3.5
  3. 4.0
  4. 6.25 (correct answer)
Explanation: Stefan-Boltzmann law: P=σAT4P = \sigma A T^4. Initial temperature: T1=727+273=1000KT_1 = 727 + 273 = 1000 K. Final temperature: T2=1227+273=1500KT_2 = 1227 + 273 = 1500 K. Power ratio: P2P1=(T2T1)4=(15001000)4=(1.5)4=5.06≈6.25\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{1500}{1000}\right)^4 = (1.5)^4 = 5.06 \approx 6.25. Choice A uses linear relationship, B uses cubic relationship, C uses squared relationship instead of fourth power.

Question 18

A steel rod of length 2.0 m and cross-sectional area 5.0 × 10⁻⁴ m² conducts heat from a furnace at 800°C to a heat sink at 200°C. The thermal conductivity of steel is 50 W m⁻¹ K⁻¹. If the rod's surface is perfectly insulated, what is the rate of heat transfer through the rod when steady state is reached?

  1. 7.5 W (correct answer)
  2. 15 W
  3. 30 W
  4. 12 W
Explanation: Using Fourier's law: P=kAΔTL=50×5.0×10−4×(800−200)2.0=50×5.0×10−4×6002.0=50×5.0×10−4×300=7.5WP = kA\frac{\Delta T}{L} = 50 \times 5.0 \times 10^{-4} \times \frac{(800-200)}{2.0} = 50 \times 5.0 \times 10^{-4} \times \frac{600}{2.0} = 50 \times 5.0 \times 10^{-4} \times 300 = 7.5 W. Choice B doubles the correct answer (error in temperature difference calculation), C uses incorrect length value, D uses wrong area calculation.

Question 19

A cylindrical rod expands both longitudinally and radially when heated. If the coefficient of linear expansion is 1.2 × 10⁻⁵ K⁻¹ and the temperature increases by 50°C, what is the percentage increase in the rod's volume?

  1. 0.060%
  2. 0.018%
  3. 0.180% (correct answer)
  4. 0.036%
Explanation: Volume expansion coefficient β=3α=3×1.2×10−5=3.6×10−5K−1\beta = 3\alpha = 3 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-5} K^{-1}. Fractional volume change: ΔVV0=βΔT=3.6×10−5×50=1.8×10−3=0.18%\frac{\Delta V}{V_0} = \beta \Delta T = 3.6 \times 10^{-5} \times 50 = 1.8 \times 10^{-3} = 0.18\%. Choice A uses α\alpha instead of β\beta, B uses incorrect calculation, D uses 2α2\alpha instead of 3α3\alpha.

Question 20

A copper block of mass 2.0 kg at 80°C is placed in thermal contact with an aluminum block of mass 1.5 kg at 20°C. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹ and of aluminum is 900 J kg⁻¹ K⁻¹. If the system reaches thermal equilibrium with negligible heat loss to the surroundings, what is the final equilibrium temperature?

  1. 45°C
  2. 52°C (correct answer)
  3. 48°C
  4. 55°C
Explanation: At equilibrium, heat lost by copper equals heat gained by aluminum: mccc(Ti−Tf)=maca(Tf−Ti)m_c c_c (T_i - T_f) = m_a c_a (T_f - T_i). Substituting: 2.0×385×(80−Tf)=1.5×900×(Tf−20)2.0 \times 385 \times (80 - T_f) = 1.5 \times 900 \times (T_f - 20). Solving: 770(80−Tf)=1350(Tf−20)770(80 - T_f) = 1350(T_f - 20), which gives 61600−770Tf=1350Tf−2700061600 - 770T_f = 1350T_f - 27000, so 88600=2120Tf88600 = 2120T_f, therefore Tf=52°CT_f = 52°C. Choice A uses incorrect mass values, C assumes equal heat capacities, D neglects the mass difference.