A refrigerator converts of water at into ice at in

PHYSICS · Single correct · CBSE · Physics · Thermal Properties of Matter · Thermal Properties of Matter

Problem

A refrigerator converts 100 g100\ \text{g} of water at 25∘C25^\circ\text{C} into ice at −10∘C-10^\circ\text{C} in 1 hour 50 minutes1\ \text{hour}\ 50\ \text{minutes}. The quantity of heat removed per minute is: (specific heat of ice =0.5 cal g−1 ∘C−1=0.5\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}, latent heat of fusion =80 cal g−1=80\ \text{cal g}^{-1}) (a) 50 cal50\ \text{cal} (b) 100 cal100\ \text{cal} (c) 200 cal200\ \text{cal} (d) 75 cal75\ \text{cal}

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