Question

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 ℃ to 15.0 ℃ I 100 *. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity of the liquid.

Answer 1

Power of heater P = 600 W

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 − 10)oC = 5oC(or 5 K)

Time taken to raise its temperature = 100s

Heat energy required to heat the liquid

ΔQ = mcΔT

And

ΔQ = P × t = 600 × 100 = 60000J

c = ΔQ/mΔT = 60000/4 x 5 = 3000 Jkg-1K-1

Heat capacity = c × m

Heat capacity = 4 × 3000JKg-1 K-1 = 1.2 × 104 J/K