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+1 vote
38.8k views
in Physics by (54.6k points)

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 s. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity of the liquid.

1 Answer

+2 votes
by (89.9k points)
edited by
 
Best answer

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 × 3000 = 1.2 × 104 J/K

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