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Heat and Thermodynamics

Section: Physics

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Question : 7 of 78

Marks: +1, -0

The temperature of water of mass 100 g is raised from

24^{\circ}\,\mathrm{C}

90^{\circ}\,\mathrm{C}

by adding steam to it. The mass of the steam added is

(Latent heat of steam

=540\,\mathrm{cal}\,\mathrm{g}^{-1}

and specific heat capacity of water

=1\,\mathrm{cal}\,\mathrm{g}^{-1}\,{}^{\circ}\mathrm{C}^{-1}

[AP EAPCET 24 May 2025 S1]

$\;10\,\mathrm{g}$
$\;12\,\mathrm{g}$
$\;8\,\mathrm{g}$
$\;16\,\mathrm{g}$

Solution:

Let

m=

mass of steam condensed. Then,

m

grams of steam forms

m

grams of water at

100^{\circ}\,\mathrm{C}

which further losses heat and comes at

90^{\circ}\,\mathrm{C}

.

Final mixture is water at

90^{\circ}\,\mathrm{C}

Now using Heat lost = Heat gained, we gel

Heat lost by steam in condensation + Heat lost by water formed at

100^{\circ}\,\mathrm{C}

=

Heat gained by water at

24^{\circ}\,\mathrm{C}

\; \Rightarrow m L + m c(100-90) = 100(c)(90-24)

\; \Rightarrow m(540 + 1 \times 10) = 100 \times 1 \times 66

\; \Rightarrow m = \; \frac{100 \times 66}{550} = \; \frac{10 \times 6}{5} = 12\,\mathrm{g}

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