Calculate the amount of heat that must be absorbed by `5g` of ice at `120^(@)C` to convert it to water at `2^(@)C`. Use the specfic heat of ice `(2.09 Jg^(-1) 1^(@)C^(-1))`, enthalpy of fusion of ice `(334 J g^(-1))`, and specific heat of water `(4.18 J g^(-1) .^(@)C^(-1))` for calculation
Strategu : We must calculate the amount of heat absorbed during three steps. ltbrrgt (i) Wariming `50 g` of ice from `-12^(@)C` to its melting point `0^(@)C`. This requires the specific heat of ice.
Melting the ice with no change in temperature . This needs the entholpy of fusion of ice at `0^(@)C`.
Warming the resulting liquid from `0^(@)C` to `20^(@)C`. This requires specific heat of water
`underset("Ice at" - 12^(@)C)(50 g) overset(Step 1)rarr underset("Ice at" 0^(@)C)(50 g) overset(Setp 2)rarr underset("Water at" 0^(@)C)(50g) overset(Setp 3)rarr underset("Water at" 20^(@)C)(50 g)`

Explanation

Steip`1 : q = m Delta T`
`= (50 g) ((2.90 J)/(g .^(@)C)) [0^(@)C - (-12^(@)C)]`
`1.25 xx 10^(3) J`
Step `2 : q = m Delta_(fus) H^(@)`
`= (50 g) (334 J g^(-1))`
`= 1.67 xx 10^(4) J`
Setp `3 : q = mc Delta T`
`= (50 g) ((4.18 J)/(g .^(@)C)) (20^(@)C - 0^(@)C)`
`4.18 xx 10^(4) J`
`= 22.4 kJ`