A 1600 Watt heater is used to melt a 5 kg block of ice that is initially at − 15 ∘ C . Determine the amount of time needed for the heater to convert the ice to liquid water assuming that 80 % of the energy from the heat is transferred to the ice.

Given, p = 1600 KW

Mass = 5kg

Initial temp T1 = -15°C

Final temp T2 = 0° C

Efficiency is 0.80

h1 is the specific enthalpy of ice at temperature (-364.18 kJ/kg)

h2 is the specific enthalpy of saturated water at temperature 2 (0 kJ/kg)

Q is the heat required to melt the ice

Let us determine the heat required to melt ice from -15 °C to liquid water at zero degrees Celsius.

We know that Q = [ h2 - h1 ]

(5 kg)[0 - (-368.18 kJ/kg)] 1821 Kilo Joules

We know that Process efficiency is equal to the heat rate divided by the heater power.

E = (Q/T)/P = Q/tP

Solving for time :

t = Q/EP

1821/(0.80)(1.600kJ) = 1423 S 1 min = 60 s So 27.3 mins