A refrigerator converts `100` g 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 = `0.5 "cal"//g^(@)C`, latent heat of fusion = `80 "cal"//g`)

Options

A. `50` cal
B. `100`cal
C. `200`cal
D. `75`cal

Answer

B

Explanation

Heat removed in cooling water from `25^(@)C` to `0^(@)C = 100 xx 1 xx 25 = 2500"cal"` lthrgt Heat removed in converting water into ice at `0^(@)C` = `100 xx80 = 8000 "cal"`
Heat removed in cooling ice from `0^(@)C` to `-15^(@)C = 100 xx 0.5 xx 10 = 500 "cal"`
Total heat removed in `1` hr `50` min
= `2500 + 8000 + 500 = 11000"cal"`
Heat removed per minute = `(11000)/(110) = 100 "cal"//"min"`