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Question

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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Solution

Heat gained by ice

$={\mathrm{mS}}_{\text{ice}}\times 10+{\mathrm{mS}}_{\mathrm{m}}\times 20\times {\mathrm{mL}}_{\mathrm{f}}=105\mathrm{m}.\mathrm{cal}$

Heat lost by calorimeter $=10\times 1\times 20=200\mathrm{cal}$

Heat lost by liquid$=-50{\int}_{40}^{20}\u200a\left(1+\frac{\mathrm{\theta}}{500}\right)\mathrm{d\theta}$

$\begin{array}{l}=50{\left[\mathrm{\theta}+\frac{{\mathrm{\theta}}^{2}}{1000}\right]}_{20}^{40}=50\left[\left(40+\frac{1600}{1000}\right)-\left(20+\frac{400}{1000}\right)\right]\\ =50\times 21.2=1060\mathrm{cal}\\ \therefore 105\mathrm{m}=1060+200\Rightarrow \mathrm{m}=12\mathrm{g}\end{array}$

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Ice at 0^{o}C is added to 200 g of water initially at 70^{o}C in a vacuum flask. when 50 g of ice has been added and has all melted, the temperature of flask and contents is 40^{o}C. When a further 80 g of ice is added and has all melted, the temperature of whole becomes 10^{o}C. Neglecting heat lost to surroundings the latent heat of fusion of ice is

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