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.

The temperature of 100g of water is to be raised from `24^@C` to `90^@

1.	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.
Answer» Correct Answer - 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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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.

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