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A refrigerator converts 100 g of water a...

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"cal"`

B

`100 ` cal

C

`200` cal

D

`75` cal

Text Solution

Verified by Experts

The correct Answer is:
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"` lthrgt Heat removed in cooling ice from `0^(@)C "to" -10^(@)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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