☰ Year plan · Science

Heat, specific heat and Q = mcΔt

Week 4 · 1-3 Physical Properties of Water (B): heat and temperature, internal energy, the Kelvin scale, specific heat and Q = mcΔt with worked examples, sea breeze, latent heat of vaporisation, effects on marine organisms · 3 lessons
2

Specific heat and the heat formula Q = mcΔt

Integrated Science Sec 1 Term 1, Unit 1: pp. 20–27
GoalExplain specific heat and use Q = m × c × Δt with masses in kilograms.
New words
specific heat (c) · the heat needed to raise the temperature of 1 kg of a substance by 1 K; unit J/kg·Kjoule (J) · the unit of heat and of energyΔt (change in temperature) · final temperature minus starting temperature, in °C or K
Explanation

Different substances need different amounts of heat to warm up. The specific heat c is the heat needed to raise 1 kg of a substance by 1 K, in J/kg·K. The higher c is, the more energy a given mass needs for the same rise, and the slower it warms or cools. Pure water has c = 4200 J/kg·K, salt water 3900, air 1005, aluminium 900, copper 390 and lead 128. The heat gained or lost is Q = m × c × Δt, with m in kilograms and Δt in °C or K. A negative Q means heat is lost. Always change grams to kilograms first: 300 g is 0.3 kg. In a mixture with no heat lost to the surroundings, the heat lost by the hot body equals the heat gained by the cold one. c depends on the substance, its state (ice and water differ) and impurities.

lead128copper390aluminium900air1005salt water3900pure water4200
Worked examples
How much heat warms 1.5 kg of water from 18 °C to 38 °C? Δt = 38 − 18 = 20 K. Q = m × c × Δt = 1.5 × 4200 × 20 = 126 000 J.
The same 126 000 J is given to 1.5 kg of copper. Then Δt = Q ÷ (m × c) = 126 000 ÷ (1.5 × 390) = 126 000 ÷ 585 ≈ 215 K. Copper has a much lower specific heat, so the same heat raises it far more than the water.
Convert units. 300 g of lead is warmed by 40 K. m = 0.3 kg, so Q = 0.3 × 128 × 40 = 1536 J. Using 300 instead of 0.3 would give an answer 1000 times too big.
Mixing, extension. 0.4 kg of aluminium at 100 °C is dropped into 1 kg of water at 20 °C. With no loss, heat lost = heat gained: 0.4 × 900 × (100 − T) = 1 × 4200 × (T − 20). That gives 36 000 − 360T = 4200T − 84 000, so 120 000 = 4560T and T ≈ 26.3 °C.
Class activity

Sand and water in the sun, with a grown-up. Put equal masses of dry sand and tap water in two identical cups and leave them side by side in sunlight for an hour. Touch each (only lightly) and, if you have a digital kitchen thermometer, read both. Describe which feels warmer and then which cools faster in the shade. Link what you find to the specific heat values in the table.

Practice
1
Find the heat needed to warm 0.5 kg of salt water by 10 K.
2
2 kg of aluminium is heated from 20 °C to 45 °C. Find Q.
3
12 600 J of heat is given to 0.5 kg of water. By how much does it warm?
4
Equal masses of copper and water get the same heat. Which warms more, and why?