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)`

Calculate the amount of heat that must be absorbed by `5g` of ice at `

1.	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)`
Answer» 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`

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