Find the amount of heat enegy required to convert 100 g of ice at `-10^(@)C` into stea at `120^(@)C`.
(Take `S_(ice)=0.5" cal "g^(-1).^(@)C^(-1),S_(W)=1" cal "g^(-1)^(@)C^(-1),S_("Steam")=0.5" cal "g^(-1).^(@)C,L_(f)=80" cal "g^(-1),L_(V)=540" cal "g^(-1)`)

Explanation

Mass of ice = 100 g. Temperature of ice, `t_(1)=-100^(@)C`
`S_("ice")=0.5calg^(-1)""^(@)C^(-1):L_(f)=80calg^(-1)`
`S_(w)=1calg^(-1)""^(@)C^(-1):L_(v)=540calg^(-1)`
`S_("steam")=0.5calg^(-1)""^(@)C^(-1):` Amount of heat supplied to increase the temperature of ice from `-10^(@)C` to `0^(@)C` is `Q=MS_("ice")(Deltat)`.
`Q_(1)=100xx0.5xx[0-(-10)]=100xx0.5xx10=500` cal
The amount of heat supplied to convert ice into water at `0^(@)C` is, `Q_(2)=ML_(F)=100xx80=8000cal` The Amount of heat supplied to increase the temperature of water from `0^(@)C` is `100^(@)C` is
`Q_(3)=MS_(w)Deltat=100xx1xx(100-0)=100xx100=10,000` cal
The Amount of heat supplied to convert water into steam at `100^(@)C` is,
`Q_(4)=ML_(v)=100xx540=54,000` cal
The Amount of heat supplied to increase the temperature of steam from `100^(@)C` to `120^(@)C` is
`Q_(5)=MS_("steam")(Deltat)`
`100xx0.5xx(120-100)=100xx0.5xx20=1000` cal
Total heat energy supplied, `Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)+Q_(5)=500+8000+10,000+54,000+1000=73,500` cal