Home
Class 11
PHYSICS
Find the change in the entropy in the fo...

Find the change in the entropy in the following process 100 g of ice at `0^@C` melts when dropped in a bucket of water at `50^@C` (Assume temperature of water does not change)

A

-4.5 cal/K

B

+ 4.5 cal/K

C

+5.4 cal/K

D

-5.4 cal/K

Text Solution

Verified by Experts

The correct Answer is:
B
Entropy `S_1= (DeltaQ)/(T)= (mL)/(T) = (80xx100)/(273)= (8xx10^3)/(273)` cal/K
Loss of entropy = `S_2= -(DeltaQ)/(T)= - (mL)/(T)`
`= -(80 xx 100)/((273+50))= -(8xx10^3)/(323)` cal/K
Total change of entropy
=`(8xx10^3)/(273)-(8xx10^3)/(323)` = +4.5 cal / K.
Promotional Banner

Similar Questions

Explore conceptually related problems

10 g of ice at 0^@C is mixed with 100 g of water at 50^@C . What is the resultant temperature of mixture

5 g of ice at 0^(@)C is dropped in a beaker containing 20 g of water at 40^(@)C . The final temperature will be

A piece of ice at 0^@C is dropped into water at 0^@C . Then ice will

If 2g ice at 0^(0)C is dropped into 10g water at 80^(0)C then temperature of mixture is

100 g of ice at 0^(@)C is mixed with 100 of water 80^(@)C . The final temperature of the mixture will be

If 10 g of the ice at 0^@C is mixed with 10g of water at 100^@ C , then the final temperature of the mixture will be

100 gram of ice at 0^@C is converted into water vapour at 100^@C Calculate the change in entropy.

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

Calculate the entropy change in the melting of 1 kg of ice at 0^(@)C in SI units. Heat of funsion of ice = 80 cal g^(-1) .