Question

200 g of ice at 0${}^{\circ }C$ converts into the water at 0${}^{\circ }C$ in 1 minute when heat is supplied to it at a constant rate. In how much time 200 g of water at 0${}^{\circ }C$ will change to 20${}^{\circ }C$? Specific latent heat of ice = 336 Jg${}^{-1}$​.

• A. 12 sec
• B. 15 sec
• C. 30sec
• D. 75 sec

Answer 1

The correct option is B

15 sec

Heat required to convert 200 g of ice at 0${}^{\circ }C$ to 200 g of water at

m = 200 g

L = 336 Jg${}^{-1}$​

Time = 1 min = 60 sec

0${}^{\circ }C$ = m L

= 200 $\times$ 336

= 67200 J

Heat produced in 1 second = $\frac{67200}{60}$ = 1120 J

Now, heat required to change 200 g of water at 0${}^{\circ }C$ to 200 g of water at 20${}^{\circ }C$ = m $c(t–0)=\frac{200\times 42}{100\times 20}$ = 16800 J

If 1120 J is produced in 1 second then 16800 J is produced in $\frac{16800}{1120}$ = 15 sec.