Problem

Problem

A refrigerator converts 100 g of water at 25∘C25^\circ\text{C} into ice at −10∘C-10^\circ\text{C} in one hour and 50 minutes. The quantity of heat removed per minute is: (Specific heat of ice = 0.5 cal/g∘C0.5 \text{ cal/g}^\circ\text{C}, latent heat of fusion = 80 cal/g80 \text{ cal/g}) (a) 50 cal (b) 100 cal (c) 200 cal (d) 75 cal

Browse the problem library