A thermally insulated vessel contains 150\; g of water at 0^{o}C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0^{o}C itself. The mass of evaporated water will be closest to :

(Latent heat of vaporization of water =2.10\times 10^{6}\; J\; Kg^{-1} and Latent heat of Fusion of water = 3.36\times\; 10^{5}J\; kg^{-1})
 

  • Option 1)

    150\; g

  • Option 2)

    130\; g

  • Option 3)

    35\; g

     

  • Option 4)

    20\; g

Answers (1)

Let m = mass of water evaporates in (gram)

\Delta Q_{req}=mL_{v}

So M = mass which convert into ice (in gram)

M =(150-m)

So\; \Delta Q_{req}=mL_{f}

\\ \Delta Q_{req}=\Delta Q_{rel}\\\; \; \; \; \\mL_{v}=(150-m)^{L_{f}}

m(L_{v}+L_{f})=150 L_{f}

m=\frac{150\; L_{f}}{L_{v}+L_{f}}=\frac{150\times3.36\times10^{5}}{3.36\times10^{5}+2.20\times10^{+6}}

m=20\; g


Option 1)

150\; g

Option 2)

130\; g

Option 3)

35\; g

 

Option 4)

20\; g

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