AP Physics 2 Thermodynamics — practice questions

Explore 2 free Thermodynamics questions in AP Physics 2 Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Thermodynamics questions and exercises with answers

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NEET · Physics · Thermodynamic ProcessesRead the answer and worked solution →One kilogram of ice at 0°C melts into water at 0°C. The latent heat of fusion for water is 334 kJ/kg. What is the increase in entropy of the ice during this melting?Read the answer and worked solution →

Worked practice questions

NEET · Physics · Thermodynamic Processes

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Matching pairs

Left
Isothermal
Right
T = constant
Left
Isobaric
Right
P = constant
Left
Isochoric
Right
V = constant
Left
Adiabatic
Right
Q = 0
Left
Cyclic process
Right
ΔU = 0 over one cycle
Left
Free expansion
Right
W = 0, Q = 0

Instructions

Match each thermodynamic process to its defining constraint.

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One kilogram of ice at 0°C melts into water at 0°C. The latent heat of fusion for water is 334 kJ/kg. What is the increase in entropy of the ice during this melting?

3.34 x 10^2 J/K
3.34 x 10^3 J/K
6.11 x 10^2 J/K
1.22 x 10^3 J/K

Show answer and solution

Correct answer: D. 1.22 x 10^3 J/K

1. First find the heat needed to melt the ice: Q = m x Lf = (1.00 kg)(334 kJ/kg) = 334 kJ = 3.34 x 10^5 J. 2. Convert the melting temperature to kelvins: T = 0°C + 273 = 273 K. 3. Apply the entropy definition: ΔS = Q/T = (3.34 x 10^5 J)/(273 K). 4. Carrying out the division: 3.34 x 10^5 / 273 = 1.22 x 10^3 J/K. 5. Option A, 3.34 x 10^2 J/K, and option B, 3.34 x 10^3 J/K, both just rescale the heat Q itself rather than dividing it by the temperature. 6. Option C, 6.11 x 10^2 J/K, does not follow from dividing 3.34 x 10^5 J by 273 K, so it does not match the required substitution. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy_

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