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Standard X

Physics

Latent Heat of Vaporization

4 gms of stea...

Question

4 gms of steam at $100^{0} C$ is added to 20 gms of water at $46^{0} C$ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in the container at thermal equilibrium is____.Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm- $^{0} C$ .

18 g m

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20 g m

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22 g m

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24 g m

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Solution

The correct option is C $22 g m$
Heat released by steam in conversion to water at $100^{0} C$ is
$Q_{1}$ = mL = 4 $\times$ 540 = 2160 cal.
Heat required to raise temperature of water from $46^{0} C$ to $100^{0} C$ is
$Q_{2} = m S Δ θ$ = 20 1 54 = 1080 J
$Q_{1}$ > $Q_{2}$ and $\frac{Q_{1}}{Q_{2}}$ = 2
Hence all steam is not converted to water only half steam shall be converted to water
Therefore, Final mass of water $= 20 + 2 = 22$ gm

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4 gms of steam at 1000C is added to 20 gms of water at 460C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in the container at thermal equilibrium is____.Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-0C.