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Question
Solve the following problem.
An electric kettle takes 20 minutes to heat a certain quantity of water from 0 °C to boiling point. It requires 90 minutes to turn all the water at 100 °C into steam. Find the latent heat of vaporization. (Specific heat of water = 1 cal/g °C)
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Solution
Let heat supplied by the kettle in 20 minutes be Q1 and that in 90 min. be Q2.
Using heat Q1, the temperature of the water is raised from 0 °C to 100 °C.
If the mass of water in the kettle is ‘m’ then,
`"Q"_1 = "ms"_"water" Delta"T" = "m" xx 1 xx (100 - 0)` = 100 m .....(1) .....`(because "S"_"water" = 1 "cal"//"g" °"C")`
Similarly using heat Q2 water is converted from liquid to gas,
∴ Q2 = m Lvap .…(ii)
Given that heat Q1, Q2 are supplied to water in 20 min. (t1) and 90 min (t2) respectively.
Kettle being same its conduction rate (Pcond) is same.
Using, `"P"_"cond" = "Q"_1/"t"_1 = "Q"_2/"t"_2` ...(iii)
From (i), (ii) and (iii),
`"100 m"/20 = "mL"_"vap"/90`
∴ `"L"_"vap" = 5 xx 90 = 450` cal/g
Latent heat of vaporisation for water is 450 cal/ g.
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