Question

8. Calculate the power of an electric heater required
to melt 1 kg of ice at 0°C in 30 s if the efficiency
of heater is 40%. Take specific latent heat of ice
= 336 J g-1​

Answer 1

Answer:

Given m=1 kg

Efficiency=40%

T=336

J g¯¹=336*10Jkg¯¹

If p watt is the power of heater, the energy supplied by the heater in time ts=pt joule. Since the efficiency of heater is 40 %, so 40% of the energy supplied by heater is used in melting the ice

i. e. 40/100pt=ml

or,0.40p*30=1*336*10³

or,p=336*10³/0. 40*30=84*10³ watt or 84 kW