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Ideal Gas Equation

6 gm of steam...

Question

$6 g m$ of steam at $100^{o} C$ is mixed with $6 g m$ of ice at $0^{o} C$ . The mass of steam left uncondensed

2 g m

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4 g m

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3 g m

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1 g m

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Solution

The correct option is B $4 g m$
Suppose $x$ qram steam get condensed then heat released $Q_{0} = m L = x L_{v} = S 40 x c a l$

heat need to meet $Q_{1} = m L_{f} = 6 \times 80 = 480 c a l$

heat to raise temp $Q_{2} = m S o t = 6 \times 1 \times 100 = 600 c a l$

total needed heat $= Q_{1} + Q_{2} = 1080 c a l$

but $Q_{0} = Q_{1} + Q_{2}$ [conversation of energy]

so $x = \frac{1080}{590} \Rightarrow 2 g m = x \Rightarrow$ uncondensed $= 6 - 2 = 4 g m$

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