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)`

Explanation

Given m = 1kg, efficiency = 40%, t = 30 s,
L = 336 J `g^(-1)` = 336 `xx 10^3 J kg^(-1)`
If P watt is the power of heater, then energy supplied by the heater in time t s = 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)/(100)` Pt=mL
or 0.4 P `xx` 30 = 1 `xx` 336 `xx 10^3`
or P = (336x10^3)/(0.4xx30)`
= 28 `xx 10^3` watt or 28 kW