KinexiaLEARN PHYSICS

Heat Transfer — Solar Water Heater (Worked Example)

Medium
Problem

A solar water heater warms 100 kg of water from 15°C to 45°C. How much solar energy must be captured? (c = 4186 J/(kg·K))

Given

  • m = 100 kg

  • c = 4186 J/(kg·K)

  • ΔT = 30 K

Find

The heat Q.

Principle

Q = mcΔT sizes the solar collector: this energy must arrive as sunlight.

Formula

Solution

  1. Substitute: Q=(100)(4186)(30)Q = (100)(4186)(30)

  2. Note: this is the useful heat; real collectors must capture extra to cover losses

  3. Multiply: Q=12,558,000 JQ = 12{,}558{,}000\ \mathrm{J}

1.26 × 10⁷ J (12.6 MJ)

Unit check

kg × J/(kg·K) × K = J. Dimensionally consistent: the result is an energy.

What this means physically

12.6 MJ ≈ 3.5 kWh — a few hours of good sunshine on a 2 m² panel. Solar thermal is efficient because it captures heat directly instead of converting to electricity first.

Common mistake

Ignoring losses. Real collectors lose heat to the air — the captured solar energy must exceed this 12.6 MJ.