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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

See also  45 g of water at 50o C in a beaker is cooled when 50 g of copper at 18 C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature.
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