Thermodynamics Equations Worksheet
Practice thermodynamic equations with 10 problems on heat capacity, latent heat, and the ideal gas law. Full worked solutions included.
Equations you will need
| Q = mcΔT | Heat energy = mass × specific heat capacity × temp change |
| Q = mL | Heat for phase change (latent heat) |
| pV = nRT | Ideal gas law |
| p₁V₁/T₁ = p₂V₂/T₂ | Combined gas law |
Symbol key
| Symbol | Quantity | Unit |
|---|---|---|
| Q | heat energy | J |
| m | mass | kg |
| c | specific heat capacity | J/kg·K |
| ΔT | temperature change | K or °C |
| L | specific latent heat | J/kg |
| p | pressure | Pa |
| V | volume | m³ |
| n | moles | mol |
| R | gas constant | 8.31 J/mol·K |
| T | temperature | K |
Practice problems
Attempt each problem on paper first, then click Show answer to check your working.
-
How much energy is needed to heat 2 kg of water by 30°C? (c = 4200 J/kg·K)
Show answer
Q = mcΔT = 2(4200)(30) = 252,000 J -
Find the energy to melt 0.5 kg of ice. (L_f = 334,000 J/kg)
Show answer
Q = mL = 0.5(334,000) = 167,000 J -
0.2 kg of water at 20°C is heated by 50,400 J. Find the final temperature.
Show answer
ΔT = Q/(mc) = 50400/(0.2 × 4200) = 60°C; final = 80°C -
2 moles of gas at 300 K occupy 0.05 m³. Find the pressure. (R = 8.31)
Show answer
p = nRT/V = (2)(8.31)(300)/0.05 = 99,720 Pa -
A gas at 100 kPa and 300 K is compressed to half its volume at constant temperature. Find the new pressure.
Show answer
p₁V₁ = p₂V₂ → p₂ = 200 kPa -
How much energy is needed to vaporise 0.1 kg of water at 100°C? (L_v = 2,260,000 J/kg)
Show answer
Q = mL = 0.1(2,260,000) = 226,000 J -
A 500 g aluminium block (c = 900 J/kg·K) absorbs 13,500 J. Find the temperature rise.
Show answer
ΔT = Q/(mc) = 13500/(0.5 × 900) = 30°C -
A gas at 27°C is heated at constant pressure to 127°C. Original volume 2 L. Find final volume.
Show answer
V₁/T₁ = V₂/T₂ → V₂ = 2(400/300) = 2.67 L -
Energy needed to heat 1 kg of ice from -10°C to 0°C then melt it. (c_ice=2100, L_f=334000)
Show answer
Q = (1)(2100)(10) + (1)(334000) = 21,000 + 334,000 = 355,000 J -
1 mole of gas at 1 atm (101,325 Pa) and 273 K. Find the volume. (R = 8.31)
Show answer
V = nRT/p = (1)(8.31)(273)/101325 = 0.0224 m³ (22.4 L)
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About this worksheet
This thermodynamics equations worksheet covers the essential equations for thermal physics at the A-Level / AP Physics 2 level. Every problem has been written to mirror the style and difficulty of real exam questions, with full algebraic working shown in the solutions.
If you find these problems too straightforward, try the more advanced worksheets in the same topic listed above. If they feel too difficult, start by reviewing the equation definitions in the box at the top of this page and then return to question 1.