Question

A refrigerator converts 100g of water at 25°C into ice at -10°C in
one hour and 50 minutes. The quantity of heat removed per minute
is (specific heat of ice = 0.5cal/g°C, latent heat of fusion = 80 cal / g)
(A) 50 cal
(B) 100 cal
(C) 200 cal
(D) 75 cal​

Answer 1

Answer:

B

Explanation:

first calculate the total heat lost in 110mins(1hr50min).

Then find out for 1 min..