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))

Find the amount of heat enegy required to convert 100 g of ice at -10^

1.	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))
Answer» Solution :Mass of ice=100g. TEMPERATURE of ice, `t_(1)=-100` `.^(@)C`. `S_(ice)=0.5" cal "g^(-1).^(@)C^(-1):L_(f)=80" cal "g^(-1)` `S_(W)=1" cal "g^(1).^(@)C^(-1):L_(V)=540cal" cal "g^(-1)` `S_("steam")=0.5" cal "g^(-1).^(@)C^(-1)`: amount of heat supplied to is `Q=M" "M_(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 OFHEAT supplied to increase the temperature of water from `0.^(@)C` to `100.^(@)C` is. `Q_(3)=MS_(W)Deltat=100xx1xx(100-0)`