Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.
The point is that the ice is absorbing heat from the surroundings all the time at a rate given by
P=ml/t=0.0021xl/120 soso in 2 minutes 0.0021l Joules of energy is absorbed where l=specific latent heat of ice
It also absorbs this heat when the 40W heater is switched on so we subtract the two maases to get the mass of ice melted by the heater
W=(m-0.021)l
40x120=0.0163l-0.0021l=0.0142l so l=4800/0.0142=338028J/K/Kg