The following graph represents change of state 1 gram of ice at  $-{20}^{0}C$ . Find the net heat required to convert ice into steam at ${100}^{0}C$. Given that $s_{ice}=0.53cal/g-{}^{0}C$and  L_v = 539 cal per gram       

In the figure

a to b: Temperature of ice increases until it reaches its melting point  $0^{0}C$

$Q_1=m{s}_{ice}\left[0-\left(-20\right)\right]=\left(1\right)\left(0.53\right)\left(20\right)=10.6cal$

b to c: Temperature remains constant until all the ice has melted

$Q_2=m{L}_{ice}=\left(1\right)\left(80\right)=80cal$

c to d: Temperature of water again rises until it reaches its boiling point ${100}^{0}C$

$Q_3=m{s}_{water}\left[100-0\right]=\left(1\right)\left(1.0\right)\left(100\right)=100cal$

d to e: Temperature is again constant until all the water is transformed into the vapour phase

$Q_4=m{L}_v=\left(1\right)\left(539\right)=539cal$

Thus, the net heat required to convert 1g of ice at  $-{20}^{0}C$ into steam at ${100}^{0}C$   is

$Q=Q_1+Q_2+Q_3+Q_4=729.6cal$