Question

300gm of water at ${25^o}C$ is added to $100gm$ of ice at ${0^o}C$ . Final temperature of the mixture is:
A) $- \dfrac{{{5^o}}}{3}C$
B) $- \dfrac{{{5^o}}}{2}C$
C) $- {5^o}C$
D) ${0^o}C$

Answer 1

In order to solve this question, one should be aware of the concept that there would be change in energy of ice when it melts. After mixing the ice with water it would take energy from water in order to melt, and as per the quantity of ice and water given in the question, the amount of water would not be able to melt the whole ice and some ice would not be melted.

Complete step by step complete answer:
As we know, that latent heat of fusion of ice is $79.7 Cal/gm$.
Let the final temperature be $T$ .
Now the heat energy lost by water is equal to the heat energy gained by the ice as it melts.
So, we have,
${m_1}S\Delta T = {m_2}L$
After putting the values of ${m_1}$ , $S$ , ${m_2}$ , $L$ we get,
$300 \times 1 \times (25 - T) = 100 \times 75$
Keeping $(25 - T)$ on one side and shifting the other values in the other side we get,
$(25 - T) = \dfrac{{100 \times 75}}{{300}}$
On Simplifying the Right Hand side we get,
$(25 - T) = 25$
On solving for $T$ we get,
$T = {0^o}C$
After that total energy left would be $4.7 \times 100$.
Total mass of water after mixing would be $300gm + 100gm = 400gm$
Amount of water again converted into ice would be given by,
$m = \dfrac{{470}}{{79.7}}$
On solving we get,
$m = 5.9gm$
So, here the whole mass is converted into water at ${0^o}C$ ,and $5.9gm$ of ice is left whose temperature is also ${0^o}C$.

After achieving the temperature of ${0^o}C$ , 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 ${0^o}C$ .

Note: Latent Heat of Fusion is the heat per unit mass required for ice to change its phase and turn into liquid. In the question given above the Latent Heat of Fusion for the whole ice would be equated to the change in energy of the water after the ice is added to the water.