KinexiaLEARN PHYSICS

Heat Transfer — Cooling Coffee (Worked Example)

Easy
Problem

A 0.30 kg cup of coffee (c = 4186 J/(kg·K)) cools from 85°C to 60°C. How much thermal energy does it lose?

Given

  • m = 0.30 kg

  • c = 4186 J/(kg·K)

  • ΔT = 25 K

Find

The heat lost Q.

Principle

Q = mcΔT: mass × specific heat × temperature change. Celsius degrees equal kelvin increments.

Formula

Solution

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

  2. Substitute: Q=(0.30)(4186)(25)Q = (0.30)(4186)(25)

  3. Multiply: Q=31,395 JQ = 31{,}395\ \mathrm{J}

3.14 × 10⁴ J (31.4 kJ)

Unit check

kg × J/(kg·K) × K = J — correct.

What this means physically

31.4 kJ flows into the air and cup — which is why coffee goes lukewarm in ~20 minutes. A lid slows the loss dramatically.

Common mistake

Using ΔT = 60. The change is 85 − 60 = 25 K — always subtract first.