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

Explanation

Quantity of beat required to convert ice into water at `0^(0)` C `(Q_(1)) = mL_("ice") = 10^(-2) xx 0.335 xx 10^(6) = 3350 `J
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) xx 4186 xx 100 = `4186 J
Quantity of heat required to convert water into steam at `100^(@)` C
`(Q_(3)) = mL_("steam") = 10^(-2) xx 2.26 xx 10^(6) ` = 22600 J
Total amount of heat required (Q) =`Q_(1) + Q_(2) + Q_(3)`
= 3350 + 4186 + 22600 = 30136 J.