Question: 120 g of ice at 0<sup>0</sup>C is mixed with 100 g of water at 80<sup>0</sup>C. Latent heat of fusio...

Question

120 g of ice at 0⁰C is mixed with 100 g of water at 80⁰C. Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g-⁰C. The final temperature of the mixture is –

Answer 1

Solution

Heat that can be rejected by 100 gm H₂O at 80⁰C when its temp. becomes 0⁰C is

Q = mCDT = (100)(1)(80) = 8000 cal

and it can melt m = $\frac{\theta}{L}$ = 100 gm ice only

Hence temperature of mixture is 0⁰C