$200$ gm water is in a thin glass container $A$ which again is kept in another adiabatic container $B$ containing $200$ gm of water.There is a heater of $1000$ watt in the container $B$ . Initially the temperature of the whole system is $100^{0} C$ . If the time in which 100 gm of water in container $A$ will convert into steam is $100 n$ sec, find the value of $n$ . Take Latent heat of vaporization $= 2.0 \times 10^{6} j o u l e / k g$ and boiling
point of water $= 100^{0} C$

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Solution

Heat will flow from container B to A after all the water in B is converted into steam.
Total steam formed is $= 300 g m s$
Total heat used for formation of steam is $Q = 0.3 k g \times 2 \times 10^{6} j o u l e k g^{- 1}$
Total heat supplied to the system in time t is $H = R a t e \times t = 1000 t j o u l e$
Equating both $0.3 k g \times 2 \times 10^{6} j o u l e k g^{- 1} = 1000 t j o u l e$
$∴ t = 600 s e c s = 100 n s e c s$
$∴ n = 6$

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