Question

$300$ gm of water at ${25}^o$C is added to $100$gm of ice at $0^{o}C$ . Final temperature of the mixture is:

• A. $-{\frac{5}{3}}^{o}C$
• B. $-{\frac{5}{2}}^{o}C$
• C. $-5^{o}C$
• D. $0^{o}C$

Answer 1

The correct option is D $0^{o}C$

We know that latent heat of fusion of ice is $79.7Cal$ per gram.

Let final temperature be T.

Then

$m_{1}S\Delta T=m_{2}L$

$300\times 1\times \left(25-T\right)=100\times 75$

$\left(25-T\right)=\frac{100\times 75}{300}$

$25-T=25$

$T=0^{\circ }C$

After that total energy left$=4.7\times 100$

Total mass of water$=400g$

Amount of water again converted into ice

$m=\frac{470}{79.7}$

$m=5.9g$

Thus whole mass is converted into water at $0^{\circ }C$,and about $5.9g$water is again converted into ice whose temperature is also $0^{\circ }C$.

After achieving the temperature of $0^{\circ }C$, latent heat of fusion is required firstly for conversion of water into ice then further lowering of temperature is possible. So the final temperature will be $0^{\circ }C$.