KVPY Sample Paper KVPY Stream-SX Model Paper-30

120 g of ice at \[0{}^\circ C\] is mixed with 100 g of water at \[80{}^\circ C.\] Latent heat of fusion is 80cal/g and specific heat of water is \[1\,\,cal/g{}^\circ C.\] The final temperature of the mixture is -

A) \[0{}^\circ C\]

B) \[40{}^\circ C\]

C) \[20{}^\circ C\]

D) \[10{}^\circ C\]

Correct Answer: A

Solution :

Heat rejected by 100 g of water at \[80{}^\circ C\]when its temperature becomes \[O{}^\circ C\] is

\[Q=m\,s\Delta \theta =(100)\,\,(1)\,\,(80)=8000\,cal\]

But this heat can melt \[m=\frac{Q}{L}\]

\[=\frac{8000}{80}=100g\] of ice only

So, temperature remains \[0{}^\circ C\]