IGCSE Physics 0625 · Section 2 · Thermal Physics — Numerical Solutions
Thermal Physics — Numerical Practice: Solutions
A. Specific Heat Capacity
A1
2 kg water, rise 20°C (c=4200).
✓168000 J
A2
500 g metal, 9000 J, rise 30°C. Find c.
1
Convert mass to kg
500 g = 0.5 kg
2
c = ΔE/(mΔθ)
9000 ÷ (0.5 × 30)
✓600 J/(kg°C)
A3
3 kg, c=900, rise 15°C.
✓40500 J
A4
4 kg, c=800, absorbs 24000 J. Find rise.
1
Δθ = ΔE/(mc)
24000 ÷ (4 × 800)
✓7.5°C
A5
1.5 kg, c=2500, cools 80→20°C. Find energy released.
1
Δθ = 80 − 20 = 60°C
ΔE = 1.5 × 2500 × 60
✓225000 J
A6
Compare 1 kg water vs 1 kg aluminium, 10°C rise.
1
Water
1 × 4200 × 10 = 42000 J
2
Aluminium
1 × 900 × 10 = 9000 J
✓Water needs 33000 J more than aluminium (42000 − 9000)
A7
Describe an experiment to measure specific heat capacity of a metal block.
✓Insert an electrical heater and a thermometer into holes drilled in the metal block. Record the starting temperature. Switch on the heater for a measured time, noting the current and voltage to calculate the electrical energy supplied (E = IVt). Record the final temperature and calculate the temperature rise. Then use c = ΔE/(mΔθ), using the known mass of the block.
A8
2 kg, 200 W heater for 60 s, c=1000. Find energy, then rise.
1
Energy = power × time
200 × 60 = 12000 J
2
Δθ = ΔE/(mc)
12000 ÷ (2 × 1000)
✓Energy = 12000 J; temperature rise = 6°C
B. Temperature Conversion
B1
Convert 25°C to K.
1
T(K) = θ(°C) + 273
25 + 273
✓298 K
B2
Convert 373 K to °C.
1
θ(°C) = T(K) − 273
373 − 273
✓100°C
B3
Convert −18°C to K.
1
T(K) = θ(°C) + 273
−18 + 273
✓255 K
B4
Convert 0 K to °C; explain physically.
1
θ(°C) = T(K) − 273
0 − 273
✓−273°C — this is absolute zero, the lowest possible temperature, at which particles have the least possible kinetic energy
C. Gas Behaviour (pV = constant)
C1
300 kPa, 0.4 m³ → 0.1 m³. Find new pressure.
1
p₁V₁ = p₂V₂
300 × 0.4 = p₂ × 0.1
✓1200 kPa
C2
150 kPa, 0.9 m³ → 450 kPa. Find new volume.
1
p₁V₁ = p₂V₂
150 × 0.9 = 450 × V₂
✓0.3 m³
C3
2 atm, 5 L → 10 L. Find new pressure.
1
p₁V₁ = p₂V₂
2 × 5 = p₂ × 10
✓1 atm
C4
Why does compressing a gas increase pressure?
✓Reducing the volume means the gas particles have less space to move in, so they collide with the container walls more frequently (in a given time) — more frequent collisions mean a greater force per unit area, i.e. a greater pressure.
C5
100 kPa, 20 cm³ → 8 cm³. Find new pressure.
1
p₁V₁ = p₂V₂
100 × 20 = p₂ × 8
✓250 kPa
C6
Shape of p vs V graph at constant temperature.
✓A curve showing pressure decreasing as volume increases (p is inversely proportional to V) — it never touches either axis. Plotting p against 1/V instead gives a straight line through the origin.
Mastery Test — Answer Key
- 323 K (50 + 273)
- 27°C (300 − 273)
- 24000 J (4 × 500 × 12)
- 500 J/(kg°C) — c = 5000 ÷ (2×5)
- 5°C — Δθ = 18000 ÷ (3×1200)
- 100 kPa — p₁V₁=p₂V₂ → 200×0.3 = p₂×0.6
- 6 atm — p₁V₁=p₂V₂ → 3×4 = p₂×2
- Absolute zero (−273°C) is the lowest possible temperature, at which particles have the least possible kinetic energy.
- Heating increases the average kinetic energy (and speed) of the gas particles. Faster particles collide with the container walls more often and with greater force, increasing the pressure.
- 105000 J (1 × 4200 × 25)
