Question

If 10g ice at ${0}^{0} C$ is mixed with 10g water at ${20}^{0} C$, then the final temperature will be:
A) ${50}^{0} C$
B) ${10}^{0} C$
C) ${0}^{0} C$
D) ${15}^{0} C$

Answer 1

Hint
For any heat transfer problem, you need to remember 1 rule and that is, heat is always transferred from hotter body to colder body. The amount of heat lost by the hotter body will be equal to the amount of heat gained by the cooler body. Here, the heat lost by the water will be equal to the heat gained by ice to convert from ice to water and then the remaining heat will be used to further raise the temperature of the water.

Complete step by step answer
In this question we are given 10g ice at ${0}^{0} C$ and 10g water at ${20}^{0} C$, if they both are mixed, heat will flow from water to ice, i.e. hotter body to cooler body. In this process, the amount of heat lost by water will be:
$Q\, = \, - ms(T - {T_0})$
As heat is lost, we are putting a negative sign there.
Where, $m$ is the mass of water
$S$ is the specific heat of water
$T$ is the final temperature of the combination
${T_0}$is the initial temperature of the water.
This heat supplied by the water is gained by ice. This heat will be used to transform some of the ice to water and then increase its temperature. We don’t yet know how much ice is converted to liquid water. Let’s assume that $\Delta m$ of ice is converted to water. Energy gained by ice is equal to:
$Q\, = \left(
\Delta m\,L{\text{ }};if{\text{ }}\Delta m < m{\text{ }}or{\text{ }}if{\text{ }}the{\text{ }}heat{\text{ }}from{\text{ }}water{\text{ }}is{\text{ }}less{\text{ }}than{\text{ }}the{\text{ }}total{\text{ }}latent{\text{ }}heat{\text{ }}in{\text{ }}ice \\
m\,L + {\text{ }}m{\text{ }}s{\text{ }}(T - {T_0});{\text{ }}if{\text{ }}\Delta m = {\text{ }}m{\text{ }}or{\text{ }}if{\text{ }}the{\text{ }}heat{\text{ }}from{\text{ }}water{\text{ }}is{\text{ }}less{\text{ }}than{\text{ }}the{\text{ }}total{\text{ }}latent{\text{ }}heat{\text{ }}in{\text{ }}ice{\text{ }} \\
\right)$
Where L is the latent heat of fusion.
Here we see that the heat lost from water = $Q\, = \, - m{\text{ }}1{\text{ }}(T - {T_0}) = m{\text{ }}(20 - {T_{}})$. Now $T$ cannot get lower than ${0}^{0} C$. So even if we put $T = 0$, the heat lost by water=20m which is much less than the total latent heat of the ice i.e. $ = \,mL = \,80{\text{ }}m{\text{ }}$.
So the whole ice will not melt. So, the temperature will stay at ${0^o}C$.
So the correct option is (C).

Note
If you don’t pay attention while checking if the heat lost by water is sufficient to melt ice, then you will end up getting negative temperatures. Which is not possible because the min temperature is only ${0^o}C$.
The sign convention of heat transfer is such that the heat lost from a body should be taken as negative and heat gained from that body to the surroundings, should be treated as positive.