Question

How much heat is required to change 456 g of ice at -20.0Degree C into water at 25.0Degree C? specific heat of water = 4186]/(kg-K); specific heat of ice = 2090 J/(kg.K) and latent heat of fusion of water = 33.5 times 10^4 J/kg.

Answer 1

Total heat will have three component, first to rise the temperature of ice from -20 to 0, second to melt the ice to water at 0 and then raise the temperature of water from 0 to 25 hence

$Q=Q_1+Q_2+Q_3=mS_{ice}(0-(-20))+mL_{fus}+mS_{wat}(25-0)\\ \Rightarrow Q=0.456\times2090\times20+0.456\times33.5\times10^4+0.456\times4186\times25=2.1954\times10^5$