KinexiaLEARN PHYSICS

Specific Heat — Heating Water for Tea (Worked Example)

Easy
Problem

How much thermal energy is needed to heat 2.0 kg of water from 20 °C to 80 °C? (Specific heat of water c = 4186 J/(kg·°C).)

Given

  • m = 2.0 kg

  • c = 4186 J/(kg·°C)

  • ΔT = 80 − 20 = 60 °C

Find

Heat energy Q.

Principle

The heat needed for a temperature change is proportional to mass, specific heat capacity, and the temperature change.

Formula

Solution

  1. Write the formula: Q=mcΔTQ = mc\Delta T

  2. Substitute: Q=2.0×4186×60Q = 2.0 \times 4186 \times 60

  3. Multiply: Q=502,320 JQ = 502{,}320\ \mathrm{J}

≈ 5.02 × 10⁵ J (≈ 502 kJ)

Unit check

kg × (J/(kg·°C)) × °C = J. Mass and temperature units cancel — correct.

What this means physically

Half a megajoule — a 1500 W kettle delivers this in about 5½ minutes (502,320/1500 ≈ 335 s). Water's huge specific heat is why it is such an effective coolant and why oceans moderate climate.

Common mistake

Using the final temperature (80 °C) instead of the change (60 °C), or converting Celsius to kelvin unnecessarily — for ΔT, a degree is a degree.