10 gm of ice at $- 200^{0} C$ is mixed with 4 gm of steam at $600^{0} C$ in an adiabatic vessel. The latent heat of vapourization of water is L cal/gm. Assume specific heats of ice, water and steam as $L / 1000, L / 500 a n d L / 1000$ in $\frac{c a l}{g m^{0} C}$ respectively and latent heat of ice as $\frac{L}{6} \frac{c a l}{g m}$ . In thermal equilibrium

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The temperature of the mixture is $100^{0} C$

The temperature of the mixture is less than $100^{0} C$

The mixture has 0.33 g of steam and 13.67 g of water

The mixture has only steam

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let the system be water at $100^{0} C$
Heat released by steam $= (4) (\frac{L}{1000}) (500) + 4 L = 6 L$
Heat required by ice $= (10) (\frac{L}{1000}) (200) + 10 (\frac{L}{6}) + 10 (\frac{L}{500}) (100)$
$= 5.67 L$
Using 0.33 L of available heat 0.33g converts into steam.