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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
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B
50oC
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C
40oC
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D
100oC
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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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