Question

A refrigeratior converts `50` gram of water at `15^(@)C` intoice at `0^(@)C` in one hour. Calculate the quantity of heat removed per minute. Take specific heat of water `=1` cal `g^(-1) .^(@)C^(-1)` and latent heat of ice `=80` cal `g^(-1)`

Answer 1

Correct Answer - `[79.2 cal. mi n^(-1)]`
(i) Heat drawn in cooling water from `15^(@)C` to `0^(@)C`,
`Delta Q_(1)=s m Delta T= 1xx50xx15=750` cal
(ii) Heat drawn in coverting `50` gram water at `0^(@)C` to ice at `0^(@)C`,
`Delta Q_(2)=m L=50xx80=4000` cals
Total heat drawn,
`Delta Q=Delta Q_(1)+DeltaQ_(2)=750+4000=4750` cal
Heat removed//minute `=(4750)/(60)=79.2 cal. mi n^(-1)`