(a) How much heat should be provided to ice of 720 g mass, lying at -`10^@C` to melt it to water at `0^@C` ? (b) How much heat should be provided to water at `0^@C` to increase its temperature to `100^@C` ?(c) How much heat should be given to water at `100^@C` to transform it completely into water ?
(d) Totally, how much heat should be given to ice of 720 g at `-10^@C` to convert it completely into vapour ? (C"ice"=2220 "J kg"^(-1) K^(-1), C"water"=4190 "J kg"^(-1) K^(-1) , LF=333 "kJ"/"kg", LV=2256 "kJ"/"kg"`)

Explanation

(a) The temperature of ice will not increase until it melts completely. Thus the heat to be given to ice to carry its temperature from `T_i=-10^@C` to `T= 0^@C` (thereafter it will start to melt) is `Q_1=C_"ice"m(T_f-T_i)`
where,
`C_"ice"`=Specific heat of ice at `-10^@C`
`=2220 "J"/"KgK"`
`therefore Q_1=2220xx0.720 xx[0-(-10)]`
=15984 J
`therefore Q_1`= 15.98 kJ
Until the ice melts completely, its temperature does not increase above `0^@ C`. Hence, the heat to be given to ice melt it completely is
`Q_2=L_F m=(333 kJ//kg)(0.720 kg)`
`thereforeQ_2`=239.8 kJ
`therefore` Required total heat =`Q_1+Q_2`
=15.98 + 239.8
=255.78 kJ
(b)Now to increase the temperature of 0.720 kg water from `T_i = 0^@C` to `T_f = 100^@C`, the required amount of heat is
`Q_3=C_"water" m[T_f-T_i]`
=4190 x 0.720 x (100-0)
`Q_3`=301680 J
`therefore Q_3` = 301.68 kJ
(c) The heat to be given to water at `100^@C` to transform it completely into vapour is
`Q_4=L_V m=2256xx0.720`
=1624.32 kJ
(d)The total amount of heat to be given to 720 g of ice at `-10^@C` to transform it completely into vapour is
`Q=Q_1+Q_2+Q_3+Q_4`
=15.98+239.8+301.68+1624.32
=2181.78 kJ