Question

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10:0°C to 15-0 °C in 100 s. Calculate : (i) the heat capacity of 40 kg of liquid, and (ii) the specific heat capacity of liquids​

Answer 1

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 × 3000JKg-1 K-1 = 1.2 × 104 J/K/151360/an-electric-heater-of-power-600-w-raises-the-temperature-of-4-0-kg-of-a-liquid-from