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The temperature of 100g of water is to b...

The temperature of 100g of water is to be raised from `24^@C` to `90^@C` by adding steam to it. Calculate the mass of the steam required for this purpose.

Text Solution

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The correct Answer is:
A, B
Let m be the mass of the steam required to raise
the temperature of 100 g of water from `24^@C` to
`90^@C`.
Heat lost by steam = Heat gained by water.
`:. m(L + sDeltatheta_1) = 100sDeltatheta_2`
or `m=((100)(s)(Deltatheta_2))/(L + s(Deltatheta_1))`
Here, s = specific heat of water `= 1 cal//g-^@C`,
L = latent heat of vaporization `=540 cal//g`
`Deltatheta_1 = (100-90) = 10^@C`
and `Deltatheta_2 = (90-24) = 66^@C`
Substituting the values, we have
`m=((100)(1)(66))/((540) + (1)(10)) = 12g `
`:. m = 12g` .
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