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Question: When \(M_1\) gram of ice at \(-10^{\circ}C\) (Specific heat = 0.5 \({calg^{-1}}^{\circ}C^{-1}\)) is ...

When M1M_1 gram of ice at −10∘C-10^{\circ}C (Specific heat = 0.5 calg−1∘C−1{calg^{-1}}^{\circ}C^{-1}) is added to M2M_2 gram of water at 50∘C50^{\circ}C (Specific heat = 1calg−1∘C−1{calg^{-1}}^{\circ}C^{-1}), finally no ice is left and the water is at 0∘C0^{\circ}C. The value of latent heat of ice, in calg−1calg^{-1} is:
(A). 5M1M2−50\dfrac{5M_1}{M_2}-50
(B). 5M2M1\dfrac{5M_2}{M_1}
(C). 50M2M1−5\dfrac{50M_2}{M_1}-5
(D). 5M2M1−5\dfrac{5M_2}{M_1}-5

Explanation

Solution

Hint: First, we need to find out the heat absorbed by the ice to rise its temperature to zero degrees and then the heat required to get converted to water. The heat gained by the ice will be supplied by the water and thus water will lose heat. Thus, by finding the heat gained by ice using the formula for latent heat and specific heat and the heat lost by water and equating them, we will get the required answer.

Formulae used:
Specific heat, c=QmΔTc=\dfrac{Q}{m\Delta T}
Latent heat of fusion, Lf=QmL_f =\dfrac{Q}{m}

Complete step by step answer:
Given that, M1M_1 gram of ice is present initially at −10∘C-10^{\circ}C, whose specific heat, Si=0.5calg−1∘C−1S_i =0.5{calg^{-1}}^{\circ}C^{-1}.
And, M2M_2 grams of water is at 50∘C50^{\circ}C, with the specific heat, Sw=1calg−1∘C−1S_w = 1{calg^{-1}}^{\circ}C^{-1}.
Now, according to the question, finally only water is left at 0∘C0^{\circ}C.
First the temperature of ice will rise from −10∘-10^{\circ} to 0∘C0^{\circ}C, so by using the formula for heat gain, we can find the heat absorbed by ice from the water to rise its temperature by 10∘10^{\circ}, we get H1=Si×M1×10=5M1H_1 = S_i \times M_1 \times 10= 5M_1 J ………. (i)
Similarly, the heat absorbed by the ice in getting converted to water will also be gained from the water. Suppose, the latent heat of ice is LfL_f.
So, heat absorbed by the ice, H2=M1LfH_2 = M_1 L_f J ………. (ii)
Since, the mass of water at 50∘C50^{\circ}C will give up the required heat to convert the mass of ice into water. And in meantime, the temperature of water will fall down to 0∘C0^{\circ}C.
So, heat lost by water, H3=M2×Sw×50=50M2H_3 = M_2\times S_w \times 50=50M_2 J ………. (iii)
Now, the total heat lost by the water will be equal to the heat gained by the ice to get converted to water.
So, using equations (i), (ii) and (iii), we can write that
H1+H2=H3  ⟹  50M2=5M1+M1Lf  ⟹  Lf+5=50M2M1H_1 +H_2 =H_3\implies 50M_2 = 5M_1 +M_1 L_f\implies L_f +5 = \dfrac{50M_2}{M_1}
∴Lf=50M2M1−5\therefore L_f = \dfrac{50M_2}{M_1}-5
Hence, option C is the correct answer.

Note: The final temperature of ice is 0∘0^{\circ} and is in liquid state, so for two processes the heat will be required; for the temperature to rise and for the ice to get converted into water. Chances of mistake that one may forget to add the heat gained during conversion of ice to water and thus can get absurd answers.