Question

# $$10\ grams$$ of steam at $$100^{o}C$$ is mixed with $$50\ gm$$ of ice at $$0^{o}C$$ then final temperature is

A
20oC
B
50oC
C
40oC
D
100oC

Solution

## The correct option is C $$40^{o}C$$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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