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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`)A. `50` calB. `100`calC. `200`calD. `75`cal |
Answer» Correct Answer - B 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"` |
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