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Byju's Answer

Standard VII

Physics

Hot and Cold

4 gm of steam...

Question

$4 g m$ of steam at $100^{\circ} C$ is added to $20 g m$ of water at $46^{\circ} C$ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass (in $g$ ) of water in container at thermal equilibrium is, (Latent heat of vaporisation $= 540 c a l / g$ , specific heat of water $= 1 c a l / g^{\circ} C$ .)

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Solution

We know that: $Q = m L$

Given: $m_{s} = 4 g, θ_{1} = 100^{\circ} C, m_{w} = 20 g, θ_{2} = 46^{\circ} C, L_{v} = 540 c a l / g, s_{w} = 1 c a l / g^{\circ} C$

Heat released by steam in conversion to water at $100^{\circ} C$ is

$Q_{1} = m_{s} L_{v} = 4 \times 540 = 2160 c a l \dots (i)$

Now
$Q = m s Δ θ$

Heat required to raise temperature of water from $46^{\circ} C$ to $100^{\circ} C$ is

$Q_{2} = m_{w} S_{w} (100 - 46)$
$= 20 \times 1 \times 54$
$= 1080 c a l \dots (i i)$

$Q_{1} > Q_{2}$ and $\frac{Q_{1}}{Q_{2}} = 2$

Hence, all steam is not converted to water only half steam shall be converted to water.

$∴$ Final mass of water $= 20 + 2 = 22 g$

Final answer: 22 g.

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Hot and Cold

Standard VII Physics

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4 gm of steam at 100∘C is added to 20 gm of water at 46∘C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass (in g) of water in container at thermal equilibrium is, (Latent heat of vaporisation =540 cal/g, specific heat of water =1 cal/g∘C.)