Question

# One hundred grams of ice is mixed with 10 gm of steam. Find the equilibrium temperature of the mixture.

Solution

## Given, Latent heat of ice $${{L}_{ice}}=336\ kJ/kg$$ Latent heat of steam$${{L}_{steam}}=\ 2260kJ/kg$$ Specific heat of water $${{S}_{water}}=4.18\ kJ/kg{{\ }^{o}}C$$ Mass of steam,$${{M}_{steam}}=0.01kg$$ Mass of ice, $${{M}_{ice}}=0.05\,kg$$ If final temperature is $${{T}_{f}}$$ Heat Loos from steam = heat gain by ice $${{M}_{steam}}{{L}_{steam}}+{{M}_{steam}}{{S}_{water}}(100-{{T}_{f}})={{M}_{ice}}{{L}_{ice}}+{{M}_{ice}}{{S}_{water}}({{T}_{f}}-0)$$ $${{T}_{f}}=\dfrac{{{M}_{steam}}{{L}_{steam}}-{{M}_{ice}}{{L}_{ice}}+{{M}_{steam}}{{S}_{water}}\times 100}{\left( {{M}_{ice}}+{{M}_{steam}} \right){{S}_{water}}}$$ $${{T}_{f}}=\dfrac{0.01\times 2260-0.05\times 336+0.01\times 4.2\times 100}{\left( 0.01+0.05 \right)\times 4.2}$$ $${{T}_{f}}={{39.6}^{o}}C\cong {{40}^{o}}C$$ Final temperature of mixture is $${{40}^{o}}C$$Physics

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