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Question: \(1Mg\) of ice at \({0^ \circ }C\) is mixed with \(10Mg\) of water \({10^ \circ }C\). The final temp...

1Mg1Mg of ice at 0∘C{0^ \circ }C is mixed with 10Mg10Mg of water 10∘C{10^ \circ }C. The final temperature is
(A) 8∘C{8^ \circ }C
(B) 6∘C{6^ \circ }C
(C) 2∘C{2^ \circ }C
(D) 0∘C{0^ \circ }C

Explanation

Solution

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 Tf{T_f}.
When 1Mg1Mg of the ice at 0∘C{0^ \circ }C is mixed with 10Mg10Mg of the water at 10∘C{10^ \circ }C, then due to the water which is at a higher temperature, the ice will melt so that at the end 11Mg11Mg will remain in the system. Now, we know that the energy required for changing the state is given by
Q=mLQ = mL.........................(1)
We know that the latent heat of fusion of the water is equal to 80cal/g80cal/g. Also, the mass of the ice is given to be equal to 1Mg1Mg. Therefore, we substitute m=1Mg=10−3gm = 1Mg = {10^{ - 3}}g, and L=80cal/gL = 80cal/g into (1), we get the energy required to melt the ice as
Q1=10−3×80cal{Q_1} = {10^{ - 3}} \times 80cal
⇒Q1=8×10−2cal\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 11Mg11Mg of water, due to which its temperature will get decreased to Tf{T_f}. So the heat extracted from the water as
Q2=−Q1{Q_2} = - {Q_1}
From (2)
Q2=−8×10−2cal{Q_2} = - 8 \times {10^{ - 2}}cal..............................(3)
We know that the energy required in changing the temperature is given by
Q=msΔTQ = ms\Delta T
Since the initial temperature of water is equal to 10∘C{10^ \circ }C, and the final temperature of water is equal to Tf{T_f}, so the heat extracted from the water is
Q2=ms(Tf−10){Q_2} = ms\left( {{T_f} - 10} \right)
We know that the specific heat of water is equal to 1cal/g−∘C1cal/g{ - ^ \circ }C. Also, the mass of the water is equal to 10Mg10Mg. Therefore substituting s=1cal/g−∘Cs = 1cal/g{ - ^ \circ }C and m=10Mg=10−2gm = 10Mg = {10^{ - 2}}g, we get
Q2=10−2×1(Tf−10){Q_2} = {10^{ - 2}} \times 1\left( {{T_f} - 10} \right)
Substituting (3) in the above equation, we get
−8×10−2=10−2×1(Tf−10)- 8 \times {10^{ - 2}} = {10^{ - 2}} \times 1\left( {{T_f} - 10} \right)
⇒(Tf−10)=−8\Rightarrow \left( {{T_f} - 10} \right) = - 8
Adding 1010 both sides, we finally get
Tf=2∘C{T_f} = {2^ \circ }C
Thus, the value of the final temperature is equal to 2∘C{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.