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 10.0 ^{0}C to 15.0^{0}C 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)^{0}C = 5^{0} 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 liquidc = △Q / [m△T]c = 60000 / (4 × 5)

c = 3000 J kg-1 K-1 = 3 × 10

^{3}J kg-1 K-1

Heat capacity = c × m

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

Heat capacity = 1.2 × 10

^{4}J / K