First slide

Change of state

Question

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

Easy

A

18 gm
B

20 gm
C

22 gm
D

24 gm

Solution

Heat released by steam in conversion to water a $100^{0} C is Q_{1} = mL = 4 \times 540 = 2160 cal . Heat required to raise$ $tempreature of water$
$from 46^{0} C to 100^{0} C i s Q_{2} = m s ∆ θ = 20 \times 1 \times 54 = 1080 J$
$Q_{1} > Q_{2} and \frac{Q_{1}}{Q_{2}} = 2$
Hence all steam is not converted to water only half steam shall be converted to water
Final mass of water = 20 + 2 = 22 gm

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