Q.
A refrigerator converts 100g of water at 10∘C into ice at 0∘C in one hour. What will be the quantity of heat removed per minute? Take the specific heat of water is cal g−1∘C−1 and latent heat of ice =80 cal g−1
Heat drawn in cooling from 10∘C to 0∘C ΔQ1=msΔL=1×100×10=1000 cal
Heat drawn in converting 100 gram water at 0∘C to ice at 0∘C ΔQ2=mL=100×80=8000cal
Net quantity of heat removed =8000+1000=9000 minute Heat removed =609000 =150 cal min−1