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AP Physics 2 Unit 9 practice questions

Try 10 multiple-choice questions from Unit 9 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    A heating element with power output PP transfers energy to a liquid of mass mm and specific heat cc. If the liquid's temperature increases from TiT_i to TfT_f with no heat loss, which expression represents the time tt required?

    t=mc(Tf−Ti)Pt = \frac{mc(T_f - T_i)}{P}

    t=Pmc(Tf−Ti)t = \frac{P}{mc(T_f - T_i)}

    t=mcP(Tf−Ti)t = \frac{mcP}{(T_f - T_i)}

    t=mc(Tf−Ti)Pt = mc(T_f - T_i)P

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    Answer: (A)

    Relating energy, power, and temperature change leads to the correct time expression. Since Q=mcΔTQ = mc\Delta T and Q=PtQ = Pt, equating them gives Pt=mc(Tf−Ti)Pt = mc(T_f - T_i), which solves to t=mc(Tf−Ti)Pt = \frac{mc(T_f - T_i)}{P}.

  2. Question 2

    A block of mass MM and specific heat CBC_B at temperature THT_H is placed into a liquid of mass mm and specific heat CLC_L at temperature TCT_C. If the system is isolated, which expression represents the final equilibrium temperature TfT_f?

    Tf=MCBTH+mCLTCMCB+mCLT_f = \frac{M C_B T_H + m C_L T_C}{M C_B + m C_L}

    Tf=MCBTH−mCLTCMCB−mCLT_f = \frac{M C_B T_H - m C_L T_C}{M C_B - m C_L}

    Tf=MCBTC+mCLTHMCB+mCLT_f = \frac{M C_B T_C + m C_L T_H}{M C_B + m C_L}

    Tf=MCBTH+mCLTCMCB−mCLT_f = \frac{M C_B T_H + m C_L T_C}{M C_B - m C_L}

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    Answer: (A)

    Applying conservation of energy allows for the derivation of the equilibrium temperature. The heat lost by the block (MCB(TH−Tf)M C_B (T_H - T_f)) equals the heat gained by the liquid (mCL(Tf−TC)m C_L (T_f - T_C)), which rearranges to this weighted average.

  3. Question 3

    A resistor of resistance RR carries a current II while immersed in a liquid of mass mm. If the liquid's temperature rises by ΔT\Delta T in time tt, which expression gives the specific heat cc of the liquid?

    c=I2RtmΔTc = \frac{I^2 R t}{m \Delta T}

    c=IR2tmΔTc = \frac{I R^2 t}{m \Delta T}

    c=I2RmtΔTc = \frac{I^2 R}{m t \Delta T}

    c=IRtmΔTc = \frac{I R t}{m \Delta T}

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    Answer: (A)

    Equating electrical energy to thermal energy isolates the specific heat. The electrical energy supplied is E=I2RtE = I^2 R t, and the thermal energy absorbed is Q=mcΔTQ = mc\Delta T. Setting I2Rt=mcΔTI^2 R t = mc\Delta T yields the correct expression.

  4. Question 4

    A lead bullet of mass mm traveling at speed vv strikes a target and stops completely. If half of the initial kinetic energy converts to internal energy in the bullet, which expression represents the bullet's temperature change ΔT\Delta T? (Let cc be the specific heat of lead).

    ΔT=v24c\Delta T = \frac{v^2}{4c}

    ΔT=v22c\Delta T = \frac{v^2}{2c}

    ΔT=v2c\Delta T = \frac{v^2}{c}

    ΔT=2v2c\Delta T = \frac{2v^2}{c}

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    Answer: (A)

    Converting a fraction of kinetic energy to heat yields the temperature change. Initial KE is 12mv2\frac{1}{2}mv^2. Half of this is 14mv2\frac{1}{4}mv^2. Equating to mcΔTmc\Delta T and canceling mm gives ΔT=v24c\Delta T = \frac{v^2}{4c}.

  5. Question 5

    A sample of material has density ρ\rho, volume VV, and specific heat cc. Which expression represents the energy QQ required to increase its temperature by ΔT\Delta T?

    Q=ρVcΔTQ = \rho V c \Delta T

    Q=ρcΔTVQ = \frac{\rho c \Delta T}{V}

    Q=VcΔTρQ = \frac{V c \Delta T}{\rho}

    Q=ρVc+ΔTQ = \rho V c + \Delta T

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    Answer: (A)

    Substituting density and volume for mass provides the correct energy expression. Since m=ρVm = \rho V, substituting this into the specific heat equation Q=mcΔTQ = mc\Delta T results in Q=ρVcΔTQ = \rho V c \Delta T.

  6. Question 6

    Water flows through a heater at a rate of RR (kg/s). The heater provides power PP. If the water enters at temperature TinT_{in} and has specific heat cc, which expression gives the exit temperature ToutT_{out}?

    Tout=Tin+PRcT_{out} = T_{in} + \frac{P}{R c}

    Tout=Tin+RcPT_{out} = T_{in} + \frac{R c}{P}

    Tout=Tin−PRcT_{out} = T_{in} - \frac{P}{R c}

    Tout=PRc−TinT_{out} = \frac{P}{R c} - T_{in}

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    Answer: (A)

    Relating power to mass flow rate yields the exit temperature. Power PP equals the rate of heat transfer Qt=mtcΔT=Rc(Tout−Tin)\frac{Q}{t} = \frac{m}{t} c \Delta T = R c (T_{out} - T_{in}). Solving for ToutT_{out} gives the correct expression.

  7. Question 7

    A calorimeter of mass mcm_c and specific heat ccc_c contains a liquid of mass mLm_L and specific heat cLc_L. If energy QQ is added to the system, which expression represents the common temperature change ΔT\Delta T?

    ΔT=Qmccc+mLcL\Delta T = \frac{Q}{m_c c_c + m_L c_L}

    ΔT=Qmccc−mLcL\Delta T = \frac{Q}{m_c c_c - m_L c_L}

    ΔT=Q(mccc)mLcL\Delta T = \frac{Q (m_c c_c)}{m_L c_L}

    ΔT=Q(mccc+mLcL)\Delta T = Q(m_c c_c + m_L c_L)

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    Answer: (A)

    Summing the heat capacities of the components correctly determines the temperature change. The total energy QQ is distributed between the calorimeter and liquid: Q=(mccc+mLcL)ΔTQ = (m_c c_c + m_L c_L)\Delta T.

  8. Question 8

    A waterfall has height hh. If all the change in gravitational potential energy of the water goes into increasing its internal energy, which expression gives the temperature increase ΔT\Delta T? (Let gg be acceleration due to gravity and cc be the specific heat of water).

    ΔT=ghc\Delta T = \frac{gh}{c}

    ΔT=mghc\Delta T = \frac{mgh}{c}

    ΔT=cgh\Delta T = \frac{c}{gh}

    ΔT=ghc\Delta T = ghc

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    Answer: (A)

    Equating potential energy change to thermal energy yields the temperature rise. mgh=mcΔTmgh = mc\Delta T. The mass mm cancels out, leaving ΔT=ghc\Delta T = \frac{gh}{c}.

  9. Question 9

    A solar collector with area AA absorbs solar radiation of intensity II (power per unit area) with efficiency ee. It heats a fluid of mass mm and specific heat cc. Which expression represents the rate of temperature change dTdt\frac{dT}{dt}?

    dTdt=eIAmc\frac{dT}{dt} = \frac{e I A}{m c}

    dTdt=IAemc\frac{dT}{dt} = \frac{I A}{e m c}

    dTdt=eImAc\frac{dT}{dt} = \frac{e I m}{A c}

    dTdt=mceIA\frac{dT}{dt} = \frac{m c}{e I A}

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    Answer: (A)

    Relating absorbed power to the rate of internal energy change provides the solution. Input power is Pin=eIAP_{in} = eIA. Since P=mc(dTdt)P = mc(\frac{dT}{dt}), solving for the rate gives dTdt=eIAmc\frac{dT}{dt} = \frac{eIA}{mc}.

  10. Question 10

    Two spheres, A and B, receive the same amount of heat energy QQ. Sphere A has mass mm and specific heat cc. Sphere B has mass 2m2m and specific heat c2\frac{c}{2}. Which expression correctly compares the temperature change of sphere A (ΔTA\Delta T_A) to sphere B (ΔTB\Delta T_B)?

    ΔTA=ΔTB\Delta T_A = \Delta T_B

    ΔTA=2ΔTB\Delta T_A = 2 \Delta T_B

    ΔTA=4ΔTB\Delta T_A = 4 \Delta T_B

    ΔTA=12ΔTB\Delta T_A = \frac{1}{2} \Delta T_B

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    Answer: (A)

    Calculating the temperature change for both spheres reveals they are equal. For A: ΔTA=Qmc\Delta T_A = \frac{Q}{mc}. For B: ΔTB=Q(2m)(c/2)=Qmc\Delta T_B = \frac{Q}{(2m)(c/2)} = \frac{Q}{mc}. Thus, the changes are identical.

What's on the AP Physics 2 Unit 9 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (40 minutes)
  • 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
  • 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

Questions are drawn from 1243 multiple-choice questions and 24 FRQs written for Unit 9, Thermodynamics.

Practice AP Physics 2 Unit 9 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 9 before you start

Other AP Physics 2 unit tests