Q.

10 gm of ice at 2000C is mixed with 4 gm of steam at  6000C 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 and L/1000 in  calgm0C respectively and latent heat of ice as  L6 calgm. In thermal equilibrium 

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

a

The temperature of the mixture is 1000C

b

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

c

The mixture has only steam 

d

The temperature of the mixture is less than 1000C

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

detailed_solution_thumbnail

Let the system be water at  1000C
Heat released by steam  =(4)(L1000)(500)+4L=6L
Heat required by ice =(10)(L1000)(200)+10(L6)+10(L500)(100)
                                     =5.67L
Using 0.33 L of available heat 0.33g converts into steam.

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon