Question

$1Mg$ of ice at ${0^ \circ }C$ is mixed with $10Mg$ of water ${10^ \circ }C$. The final temperature is
(A) \[{8^ \circ }C\]
(B) ${6^ \circ }C$
(C) \[{2^ \circ }C\]
(D) \[{0^ \circ }C\]

Answer 1

Hint: To solve this question, we need to equate the heat extracted from the water to lower its temperature, with the heat energy required to melt the ice into water. The heat energy used for changing the temperature is found by using the value of the specific heat, while the heat energy used for changing of state is found by using the value of the latent heat.

Complete step-by-step solution:
Let the final temperature be ${T_f}$.
When $1Mg$ of the ice at ${0^ \circ }C$ is mixed with $10Mg$ of the water at ${10^ \circ }C$, then due to the water which is at a higher temperature, the ice will melt so that at the end $11Mg$ will remain in the system. Now, we know that the energy required for changing the state is given by
$Q = mL$.........................(1)
We know that the latent heat of fusion of the water is equal to $80cal/g$. Also, the mass of the ice is given to be equal to $1Mg$. Therefore, we substitute $m = 1Mg = {10^{ - 3}}g$, and $L = 80cal/g$ into (1), we get the energy required to melt the ice as
${Q_1} = {10^{ - 3}} \times 80cal$
$ \Rightarrow {Q_1} = 8 \times {10^{ - 2}}cal$.....................(2)
Now, this energy is transferred from the hot water. So an equal amount of the heat is to be extracted from $11Mg$ of water, due to which its temperature will get decreased to ${T_f}$. So the heat extracted from the water as
\[{Q_2} = - {Q_1}\]
From (2)
${Q_2} = - 8 \times {10^{ - 2}}cal$..............................(3)
We know that the energy required in changing the temperature is given by
$Q = ms\Delta T$
Since the initial temperature of water is equal to ${10^ \circ }C$, and the final temperature of water is equal to ${T_f}$, so the heat extracted from the water is
${Q_2} = ms\left( {{T_f} - 10} \right)$
We know that the specific heat of water is equal to $1cal/g{ - ^ \circ }C$. Also, the mass of the water is equal to $10Mg$. Therefore substituting $s = 1cal/g{ - ^ \circ }C$ and $m = 10Mg = {10^{ - 2}}g$, we get
\[{Q_2} = {10^{ - 2}} \times 1\left( {{T_f} - 10} \right)\]
Substituting (3) in the above equation, we get
\[ - 8 \times {10^{ - 2}} = {10^{ - 2}} \times 1\left( {{T_f} - 10} \right)\]
\[ \Rightarrow \left( {{T_f} - 10} \right) = - 8\]
Adding $10$ both sides, we finally get
\[{T_f} = {2^ \circ }C\]
Thus, the value of the final temperature is equal to \[{2^ \circ }C\].

Hence, the correct answer is option C.

Note: The phenomenon of heat exchange taking place in the above question is applied in a calorimeter, and the phenomenon is known as calorimetry. A calorimeter is used in measuring the amount of heat exchange involved in a chemical reaction.