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Question

M gram of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40ºC [heat of vaporization of water is 540 cal/g and heat and heat of fusion of ice is 80 cal/ g], the value of M is_______

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Solution

$\begin{array}{l}Heatgainbyice=Heatlossbysteam\text{}\\ {M}_{ice}{L}_{f}+{M}_{ice}\left(40-0\right){C}_{w}={m}_{steam}{L}_{V}+{m}_{steam}\left(100-40\right){C}_{w}\\ \Rightarrow 200\left[80+40\left(1\right)\right]=M\left[540+60\left(1\right)\right]\text{}\\ \Rightarrow 200\left(120\right)=M\left(600\right)\\ \text{}\Rightarrow M=40g\end{array}$

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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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