Calculate the quantity of heat required to convert 10^(2) kg of ice at 0^(0)C to steam at 100^(0)C. (Given L_("ice") =0.335 xx 10^(6) JKg^(-1), L_("steam") = 2.26 xx 10^(6) JKg^(-1). Specific heat of water ="4186J Kg"^(-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^(0)C to steam at 100^(0)C. (Given L_("ice") =0.335 xx 10^(6) JKg^(-1), L_("steam") = 2.26 xx 10^(6) JKg^(-1). Specific heat of water ="4186J Kg"^(-1) K^(-1).
Answer» Solution :QUANTITY of heat required to CONVERT ice into WATER at `O^0C=, (Q_1)= mL, =10^(-2) xx 0.355 x 10^(-6)=3350J` Quantity of heat required to raise the temperature of water from `0^@C" to "100^@C`. `(Q_2) =mS(12-1) = 10^(-2)xx 4186 xx100= 4186J` Quantity of heat required to convert water into steam at `100^@C` `(Q_3)= mL_2, =10^(-2) xx 2.26 xx 100= 22600J` Total amount of heat required `(Q) =Q_1+Q_2+Q_3= 3350 + 4186 +22600= 30136J`