Two formulas cover most heat calculations: Q = mcΔT while a substance changes temperature, and Q = mL while it changes phase. Here m is mass, c is specific heat, ΔT is the temperature change, and L is latent heat, the energy absorbed or released during a phase change with no change in temperature. In the examples below the specific heat of water is 4186 J/(kg·°C).

Example 1: total heat

Heat 2 kg of water from 20°C to 50°C, ignoring heat loss to the surroundings. Q = 2 × 4186 × (50 − 20) = 251,160 J. (The earlier version stated 20 kg of water but calculated with 2 kg; this article uses 2 kg throughout.)

Example 2: thermal equilibrium

Drop a 0.5 kg copper block at 90°C into the 2 kg of water at 20°C. The specific heat of copper is 387 J/(kg·°C). The system is closed, so the heat gained by the water equals the heat lost by the copper.

2 × 4186 × (Tf − 20) = 0.5 × 387 × (90 − Tf), which gives Tf ≈ 21.58°C. Copper has a much smaller heat capacity than the water, so the water warms by only about 1.6°C.

Example 3: water to steam

Turn 2 kg of water at 20°C into steam at 100°C. The latent heat of vaporization is L = 2,260,000 J/kg.

• Heating to 100°C: 2 × 4186 × 80 = 669,760 J.
• Vaporizing: 2 × 2,260,000 = 4,520,000 J.
• Total: 669,760 + 4,520,000 = 5,189,760 J.

The heating step alone leaves the water at 100°C; more heat must be supplied for it to become steam.

Example 4: ice to steam at 160°C

Take 3 kg of ice at −40°C. Specific heats: ice 2090, water 4186, steam 2010 J/(kg·°C). Latent heats: fusion 333,000 J/kg, vaporization 2,260,000 J/kg. The calculation has five stages:

1. Ice from −40°C to 0°C: 3 × 2090 × 40 = 250,800 J
2. Melting the ice: 3 × 333,000 = 999,000 J
3. Water from 0°C to 100°C: 3 × 4186 × 100 = 1,255,800 J
4. Vaporizing the water: 3 × 2,260,000 = 6,780,000 J
5. Steam from 100°C to 160°C: 3 × 2010 × 60 = 361,800 J

The total is 9,647,400 J. The earlier version miscalculated stage 4 as 6,280,000 J and so reported 9,147,400 J; both figures have been corrected.