Your Equation

Library / Foundations: Physics / FPH-15

Foundations: Physics

Specific heat capacity

Energy to change the temperature of a mass.

Ref
FPH-15
Rev
v1
Field
Foundations: Physics
Result
E [kJ]

Use specific heat capacity c for a temperature change. E in joules, m in kg, c in J/(kg·K), ΔT in K or °C change.

The energy needed to warm something depends on its mass, how stubbornly its material resists warming (c), and the temperature rise.

Calculator

Solve for
kJ
kg
0.01100
J/(kg·K)
1005,000
K
1200
  • Energy in joules83,600 J
  • Time at 1 kW heating83.6 s

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Thermal energy (kJ) current value 83.6 kJ
mMassKey driver
1.6716,700
ΔTTemperature change
2.09418
cSpecific heat capacity
2100
  • Water has a very high specific heat, which is why it is a good coolant and heat store.
  • The same energy warms a lighter object or low-c material faster.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kJ)Local stepChange in result
m Mass0.01 to 100 kg1.67216,72016,720+10%+10 %
ΔT Temperature change1 to 200 K2.09418415.9+10%+10 %
c Specific heat capacity100 to 5,000 J/(kg·K)210098+10%+10 %

Sweep of Temperature change

Temperature change (K)Thermal energy (kJ)
12.09
1.9394.053
3.7617.86
7.29315.24
14.1429.56
27.4257.32
53.18111.2
103.1215.5
200418

Sweep one variable

Thermal energy (kJ)Hover, touch, or use the arrow keys on the chart
0100200300400 1520100 Temperature change (K), log scale now 83.6

Variables and how each one contributes

E

Thermal energy

Result · kJ

What the equation gives you. Everything else feeds it.

m

Mass

kg · typical 0.01 to 100

Mass heated. Directly proportional.

c

Specific heat capacity

J/(kg·K) · typical 100 to 5,000

Water 4180, aluminium 900, steel 490. Directly proportional.

ΔT

Temperature change

K · typical 1 to 200

Rise in temperature. Directly proportional.

Worked example with the values above

  1. m × c = 0.5 × 4,180 = 2,090 J/K.
  2. Multiply by ΔT = 40 K: 83,600 J.
  3. E = 83.6 kJ.

Assumptions and limits

  • No change of state.
  • Constant specific heat over the temperature range.

Common mistakes

  • Using grams with J/(kg·K).
  • Ignoring heat lost.

Statistics and uncertainty

Heat losses to the surroundings make measured energy higher than this ideal value, often by 10% or more.

Where it is used

  • Heating and cooling duty estimates.
  • Thermal soak of equipment.

Maths you need

  • Temperature and heat: difference between them
  • Units: J, kg, K

Related formulas

Practise this formula