A refrigerator converts `100 g` of water at `25^(circ) C` into ice at `-10^(circ) C` in `1 h` and `50 min`. The quantity of heat (in calorie) removed per min is (specific heat of ice `=0.5` cal `(g)^(-1 circ) C^(-1)`, specific heat of water `=1 (cal) (g)^(-1)/ ^circC^(-1)`, latent. heat of fusion `-80 (cal) / (g)` )

Answer

100