. An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 0C to 15.0 0C in 100 s. Calculate


An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 0C to 15.0 0C in 100 s. Calculate:

(i) the heat capacity of 4.0 kg of liquid, and

(ii) the specific heat capacity of liquid

 

Best Answer

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 and

△Q = P × t

△Q = 600 × 100

△Q = 60000 J

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

Heat capacity = 1.2 × 104 J / K

 

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