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Question: 1 kg of Ice at \( - 20^\circ {\text{C}}\) is mixed with 2 kg of water at \(90^\circ {\text{C}}\). As...

1 kg of Ice at −20∘C - 20^\circ {\text{C}} is mixed with 2 kg of water at 90∘C90^\circ {\text{C}}. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume latent heat of ice = 334.4 KJ/Kg, specific heat of water and ice are 4.18 KJ/ (Kg. K) and 2.09 KJ/ (Kg. K), respectively.)
A. 30∘C30^\circ {\text{C}}
B. 0∘C0^\circ {\text{C}}
C. 80∘C80^\circ {\text{C}}
D. 45∘C45^\circ {\text{C}}

Explanation

Solution

Hint: To solve this question, we will use the formula Q = msΔt{\text{Q = ms}}\Delta {\text{t}} and Q = mL to find the heat loss and heat gain during the process of mixing and thus find the final temperature of the mixture.

Complete step-by-step solution -
Now, the ice at −20∘C - 20^\circ {\text{C}} on mixing with water first comes to 0∘C0^\circ {\text{C}} . Then, at 0∘C0^\circ {\text{C}}it changes its state i.e. ice changes into water. Now, from −20∘C - 20^\circ {\text{C}} to 0∘C0^\circ {\text{C}} heat is absorbed by the ice. So, heat absorbed = Q = msΔt{\text{Q = ms}}\Delta {\text{t}}, where m is the mass of ice, s is the specific heat of ice and Δt\Delta {\text{t}} is the change in temperature. Also, at 0∘C0^\circ {\text{C}}, ice converts into water. So, heat is also absorbed and is equal to Q = mL, where L is the latent heat of ice.
So, we get total heat absorbed = msΔt + mL{\text{ms}}\Delta {\text{t + mL}} = 1(2.09)(20) + 1(334.4){\text{1(2}}{\text{.09)(20) + 1(334}}{\text{.4)}} = 376.2 KJ
Therefore, total heat absorbed = 376.2 KJ
Now, let the water at 90∘C90^\circ {\text{C}} lose heat and come down to 0∘C0^\circ {\text{C}}. Therefore, heat loss = Q = msΔt{\text{Q = ms}}\Delta {\text{t}}
So, heat loss = msΔt{\text{ms}}\Delta {\text{t}} = 2(4.18)(90)2(4.18)(90) = 752.4 KJ
So, heat remaining = heat loss – heat gain = 752.4 KJ – 376.2 KJ = 376.2 KJ
Heat remaining = 376.2 KJ
Now, in the question, it is given that there is no loss of heat which means that no heat is dissipated in the procedure of mixing. So, the temperature of the mixture is not 0∘C0^\circ {\text{C}}. Let the final temperature of mixture be t∘C{\text{t}}^\circ {\text{C}}. The mass of the mixture is 3 kg. As, the mixture only contains water, therefore, specific heat of final mixture = 4.18 KJ/ (Kg. K)
Therefore, heat remaining = msΔt{\text{ms}}\Delta {\text{t}}
376.2 = 3(4.18)(t)3(4.18)({\text{t)}}
t = 376.23×4.18\dfrac{{376.2}}{{3 \times 4.18}} = 30∘C30^\circ {\text{C}}
So, the final temperature of the mixture is 30∘C30^\circ {\text{C}}. So, option (A) is correct.

Note: When we come up with such types of questions, we will let the ice and water both come at 0∘C0^\circ {\text{C}}. We will find the heat gain by ice and heat loss by water by using the formulas, Q = msΔt{\text{Q = ms}}\Delta {\text{t}} and Q = mL. Now, the mixture is at 0∘C0^\circ {\text{C}}. We will subtract the heat loss and heat gain to check whether any heat is left. If heat is remaining, then it will be used to increase the temperature of the mixture. We will find the temperature by using the formula Q = msΔt{\text{Q = ms}}\Delta {\text{t}}. If there is no heat left, then the final temperature of the mixture is 0∘C0^\circ {\text{C}}.