Question

Inan experiment on the specific heat of a metal, a 0.20 kg block of themetal at 150 °C is dropped in a copper calorimeter (of waterequivalent 0.025 kg) containing 150 cm³of water at 27 °C. The finaltemperature is 40 °C. Compute the specific heat of the metal. Ifheat losses to the surroundings are not negligible, is your answergreater or smaller than the actual value for specific heat of themetal?

Answer 1

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T₁ = 150°C

Final temperature of the metal, T₂ = 40°C

Calorimeter has water equivalent of mass, m^’ = 0.025 kg = 25 g

Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C:

150 × 1 = 150 g

Fall in the temperature of the metal:

ΔT = T₁ – T₂ = 150 – 40 = 110°C

Specific heat of water, C_w = 4.186 J/g/°K

Specific heat of the metal = C

Heat lost by the metal, θ = mCΔT … (i)

Rise in the temperature of the water and calorimeter system:

ΔT′^’ = 40 – 27 = 13°C

Heat gained by the water and calorimeter system:

Δθ′′ = m₁ C_wΔT^’

= (M + m′) C_w ΔT^’ … (ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT = (M + m^’) C_w ΔT^’

200 × C × 110 = (150 + 25) × 4.186 × 13

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.