Question

M gm of ice at 0 degree celcius is to be converted to water at 0 degree celcious. IfL is the latent heat of fusion of ice, the quantity of heat required for the above opetation would be

Answer 1

Ice→water→water→steam

First 0 degree cesious of ice will be converted into 0 degree celcious of water now 0 degree celcious of water converted into 100 degree celcious of water now 100 degree celcious of water converted into 100 degree celcious of steam this is total phenomenon.

Now come on problem

Heat required is given by

Q=M×L1+M×S×(T2-T1)+M×L2 ,Where

L1=latent heat of fusion of water 334 jule/gr.

L2=latent heat of vaporisation of water 2230 j/g

S=specific heat of water 4.186 j/g

Q=1×334+1×4.186×(100-0)+1×2230

Q=2982.6 jule or

Q=2664/4.186

Q=712.517 cal