Calorimetry Heat Transfer Calculator

How to use this calculator

1. Enter or choose mass of the hot object.
2. Enter or choose specific heat of the hot object.
3. Enter or choose starting temperature of the hot object.
4. Enter or choose mass of the water (second body).
5. Enter or choose specific heat of the water (second body).
6. The final (equilibrium) temperature updates as you type. Use Copy result or Share to keep it.

Method

T_f = (m₁c₁T₁ + (m₂c₂ + C_cal)T₂) ÷ (m₁c₁ + m₂c₂ + C_cal); Q = m₁c₁(T₁ − T_f)

m₁, c₁, T₁

mass, specific heat and starting temperature of the hot object

m₂, c₂, T₂

the same for the water or second body

C_cal

heat capacity of the calorimeter cup

Worked example

1. m₁c₁ = 0.5 × 897 = 448.5 J/K; m₂c₂ = 1 × 4184 = 4184 J/K.
2. T_f = (448.5 × 100 + 4184 × 20) ÷ (448.5 + 4184) = 128 530 ÷ 4632.5 = 27.745 °C.
3. Q = 448.5 × (100 − 27.745) = 32 406 J = 32.41 kJ.

A 0.5 kg block of aluminium at 100 °C dropped into 1 kg of water at 20 °C brings both to about 27.7 °C, transferring 32.4 kJ.

What the result means

Water’s large specific heat means it changes temperature much less than the metal. Measure T_f in the lab and you can solve the same balance backwards for an unknown specific heat.

Assumptions and limits

• The system is insulated: no heat is lost to the surroundings.
• No phase change or chemical reaction occurs.
• Specific heats are constant.
• If the result is above 100 °C or below 0 °C for water, a phase change would occur and this result is not valid.