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Question

It takes 15 minutes to rise a certain amount of water from $$0^oC$$ 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:


A
525 cal
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B
540 cal
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C
533 cal
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D
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 = $$15P$$ = $$Q_{1}$$
Heat supplied in $$1$$ hour and $$20$$ minutes = $$80P$$ = $$Q_{2}$$
$$Q_{1} = mc\Delta T = m(1)(100) = 100m$$
$$Q_{2} = mL$$ L is the latent heat of vaporization.
i.e. dividing $$Q_{1} \ and \ Q_{2}$$
$$\dfrac{15}{80} = \dfrac{100}{L}$$
i.e. $$L = 533.33$$ cal. 
Hence answer is option C.

Physics

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