An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10 C to 15 C in 100 s. Calculate: (i) the heat capacity of 4.0 kg of liquid

Calorimetry numerical Question: 7 [reference: Selina class 10 ICSE Physics]

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0⁰C to 15.0⁰C in 100 s. Calculate:
(i) the heat capacity of 4.0 kg of liquid, and
(ii) the specific heat capacity of liquid

Solution:

Power of heater P = 600 W

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 – 10)⁰C = 5⁰ C (or 5 K)

Time taken to raise its temperature = 100 s

Heat energy required to heat the liquid

△Q = mc△T ……….. [1]

△Q = P × t
△Q = 600 × 100
△Q = 60000 J …………………. [2]

The specific heat capacity of liquid c = △Q / [m△T]

c = 60000 / (4 × 5)

c = 3000 J kg-1 K-1 = 3 × 10³ J kg-1 K-1

Heat capacity = c × m

Heat capacity = 4 × 3000 J kg-1 K-1 Kg

Heat capacity = 1.2 × 10⁴ J / K