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Standard X

Physics

Specific Heat

It takes 15 m...

Question

It takes 15 minutes to rise a certain amount of water from $0^{o} C$ to boiling point using a heater. Then it takes 1 hour and 20 minutes to convert all the water into vapour. Then the latent heat of vapourization of water is:

525 cal

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540 cal

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533 cal

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513 cal

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Solution

The correct option is C 533 cal
Let the power delivered by the heater = $P$ .
Heat supplied in $15$ minutes = $15 P$ = $Q_{1}$
Heat supplied in $1$ hour and $20$ minutes = $80 P$ = $Q_{2}$
$Q_{1} = m c Δ T = m (1) (100) = 100 m$
$Q_{2} = m L$ L is the latent heat of vaporization.
i.e. dividing $Q_{1} a n d Q_{2}$
$\frac{15}{80} = \frac{100}{L}$
i.e. $L = 533.33$ cal.
Hence answer is option C.

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It takes 15 minutes to rise a certain amount of water from 0oC to boiling point using a heater. Then it takes 1 hour and 20 minutes to convert all the water into vapour. Then the latent heat of vapourization of water is:

It takes 15 minutes to rise a certain amount of water from $0^{o} C$ to boiling point using a heater. Then it takes 1 hour and 20 minutes to convert all the water into vapour. Then the latent heat of vapourization of water is: