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A heater melts 0^@C ice in a bucket comp...

A heater melts `0^@C` ice in a bucket completely into water in `6` minutes and then evaporates all that water into steam in `47` minutes `30 sec`. If latent heat of fusion of ice is `80 cal//gram`, latent heat of steam will be (specific heat of water is `1 cal//gam-^@C`)

A

`536 Cal//gram`

B

`533.3 Cal//gram`

C

`540 Cal//gram`

D

`2.268 xx 10^(6) J//Kg`

Text Solution

Verified by Experts

The correct Answer is:
B
Let 'm' be the mass of ice in the bucket Heat given out by heater in `6 min` is `80 m` Heat given out in `47.5 min` is `100 m+mL_(v)`
`m xx 80 -- rarr 6 minutes`
`(m xx1 xx 100) + (m xx L_(v)) rarr 47.5 minutes`
:. `80 xx 47.5 = 6(100 + L_(v))`.
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