Question

In following equiation calculate value of H: 1kg ice at `-20^@C=H+1`kg water at `100^@C`, here H means heat required to change the state of substance.

Answer 1

Heat required to convert `1kg` ice at `-20^(@)C` into `1kg` water at` 100^(@)C = 1kg "ice at" -20^(@)C to 1kg "ice at" 0^(@)C "ice at" 0^(@)C +1kg "water at" 0^(@)C +1kg` water at `0^(@)C to 1 kg` water at `100^(@)C`
`= 1xx (1)/(2) xx 20 +1 xx 80 +1 xx 1xx 100 = 190 Kcal` So `H =- 190` Kcal
Negative sign indicate that `190` Kcal heat is with drawn form `1kg` water at `100^(@)C` to convert into `1kg` at `-20^(@)C`