. 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

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