Calculate the quantity of heat required to convert 10^(-2)kg of ice at 0^(@)C to steam at 100^(@)C. (Given L_("ice")=0.335xx10^(6)Jkg^(-1),L_("steam")=2.26xx10^(6)Jkg^(-1), Specific heat of water = 4186Jkg^(-1)K^(-1)).

Calculate the quantity of heat required to convert 10^(-2)kg of ice at

1.	Calculate the quantity of heat required to convert 10^(-2)kg of ice at 0^(@)C to steam at 100^(@)C. (Given L_("ice")=0.335xx10^(6)Jkg^(-1),L_("steam")=2.26xx10^(6)Jkg^(-1), Specific heat of water = 4186Jkg^(-1)K^(-1)).
Answer» SOLUTION :Quantity of HEAT required to CONVERT ice into water at `0^(@)C(Q_(1))=mL_("ice")=10^(-2)xx0.335xx10^(6)=3350J` Quantity of heat required to raise the temperature of water from `0^(@)C" to "100^(@)C`. `(Q_(2))=mS_(W)(t_(2)-t_(1))=10^(-2)xx4186xx100=4186J` Quantity of heat required to convert water into steam at `100^(@)C` `(Q_(3))=mL_("steam")=10^(-2)xx2.26xx10^(6)=22600J` Total amont of heat required `(Q)=Q_(1)+Q_(2)+Q_(3)` = 3350 + 4186 + 22600 = 30136 J.