Internal Energy and Temperature

Science — Second Preparatory

Internal Energy and Temperature

Worked examples

Key points

Example 1How much heat is needed to heat 2 kg of water from 20 °C to 45 °C? Take c for water = 4200 J/(kg·°C).

SolutionΔT = 45 − 20 = 25 °C

Q = 2 × 4200 × 25 = 210000 J = 210 kJ

Example 2A piece of iron of mass 0.5 kg is heated from 20 °C to 120 °C. Calculate the heat gained, given c for iron = 450 J/(kg·°C).

SolutionΔT = 120 − 20 = 100 °C

Q = 0.5 × 450 × 100 = 22500 J = 22.5 kJ

The iron needs much less heat than water to warm the same mass by the same amount, because its specific heat capacity is smaller.
Example 3 Convert 27 °C to kelvin: T = 27 + 273 = 300 K. If the room temperature rises from 27 °C to 127 °C, its kelvin value changes from 300 K to 400 K.

Simple experiment→Mix a cup of hot water with a cup of cold water. Heat moves from the hot water to the cold water until the two temperatures are nearly equal. This shows that heat always moves from the body at the higher temperature to the body at the lower temperature.

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