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Specific heat capacity

Heat and matter · level 4, lesson 3 of 6

A question to start

On a sunny day the sand on a beach gets too hot to walk on, but the sea stays cool. Both get the same sunshine. Why is the sand so much hotter?

The idea

Different materials need different amounts of energy to warm up. The energy \(Q\) needed to warm a mass \(m\) by \(\Delta T\) is

\[Q = mc\,\Delta T,\]

where \(c\) is the specific heat capacity: the energy to warm 1 kg by 1 °C (or 1 K).

Material \(c\), J per kg per °C
Water 4,180
Ice 2,100
Air 1,005
Carbon dioxide (most of Mars's air) 840
Sand, dry soil about 800
Aluminium 900
Steel 450
Copper 385

Water's value is huge. It takes about five times as much energy to warm water as to warm the same mass of sand. So the sea warms slowly, and cools slowly too.

watericeairaluminiumcarbon dioxide sandsteelcopper

Worked examples

1. Warming water. 2 kg of water from 20 °C to 70 °C: \(Q = 2 \times 4180 \times 50 = 418{,}000\) J.

2. Warming aluminium. 2 kg of aluminium by the same 50 °C: \(2 \times 900 \times 50 = 90{,}000\) J, under a quarter as much.

3. Air on Mars and Earth. A cubic metre of air at the ground:

Warming 1 m³ of outside air by 1 °C Mars Earth
Mass of the air 0.016 kg 1.217 kg
Energy needed, \(mc\Delta T\) \(0.016 \times 840 \approx 13\) J \(1.217 \times 1005 \approx 1{,}220\) J

Mars air takes about 90 times less energy to warm, and gives back 90 times less when it cools.

Try it

  1. How much energy warms 1 kg of water by 10 °C?
  2. How much warms 5 kg of steel from 20 °C to 220 °C?
  3. 4,180 J goes into 1 kg of sand. Roughly how much does it warm?
  4. Why does the sea stay cooler than the beach on a sunny day?
Answers
  1. \(1 \times 4180 \times 10 = 41{,}800\) J.
  2. \(5 \times 450 \times 200 = 450{,}000\) J.
  3. \(\Delta T = 4180/(1 \times 800) \approx 5\) °C.
  4. Water needs about five times as much energy as sand to warm by each degree.

Common mistake

Using the final temperature instead of the change. \(\Delta T\) is the rise (or fall): from 20 °C to 70 °C, \(\Delta T = 50\) °C, not 70 °C.

Where it leads