KinexiaLEARN PHYSICS

Heat Transfer — Engine Block (Worked Example)

Easy
Problem

A 5.0 kg aluminium engine part (c = 900 J/(kg·K)) heats from 20°C to 120°C. How much thermal energy does it absorb?

Given

  • m = 5.0 kg

  • c = 900 J/(kg·K)

  • ΔT = 100 K

Find

The heat absorbed Q.

Principle

Q = mcΔT with aluminium's specific heat — much lower than water's.

Formula

Solution

  1. Substitute: Q=(5.0)(900)(100)Q = (5.0)(900)(100)

  2. Note: ΔT=120−20=100 K\Delta T = 120 - 20 = 100\ \mathrm{K} — always subtract temperatures first

  3. Multiply: Q=450,000 JQ = 450{,}000\ \mathrm{J}

4.50 × 10⁵ J (450 kJ)

Unit check

kg × J/(kg·K) × K = J. The units collapse to joules — energy, as required.

What this means physically

450 kJ — aluminium's low c means engines heat up (and cool down) quickly, which is good for reaching operating temperature fast but demands good cooling.

Common mistake

Using water's c = 4186. Specific heat is material-specific — aluminium is 900, less than a quarter of water's.