The following graph represents change of state of `1` gram of ice at `-20^@C`. Find the net heat required to convert ice into steam at `100^@ C`.
`S_(ice) = 0.53 cal//g -^@ C`
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Explanation

In the figure :
`a to b` : temperature of ice increases until its reaches its melting point `0^@ C`
`Q_(1)=mS_(ice)[0-(-20)] =(1)(0.53)(20)=10.6 cal`
`b to c` : Temperature remains constabt until all the ice has melted.
`Q_(2) = mL_(f) = (1)(80) = 80 cal`
`c to d` : Temperature of water againm rises until it reaches its boiling point `100^@C`
`Q_(3) = mS_(water) [100-0] = (1)(1.0)(100) = 100 cal`
`d to e` : Temperature is again constant until all the water is transformed into the vapour phase
`Q_(4) = mL_(v) = (1)(530) = 539 cal`
Thus, the net heat required to convert `1 g` of ice at `-20^@ C` into steam at `100^@ C` is
`Q = Q_(1) + Q_(2) + Q_(3) + Q_(4) = 729.6 cal`.