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22 \mathrm{gm} of \mathrm{CO}_{2} at 27^{\circ} \mathrm{C} is mixed with 16 \mathrm{gm} of \mathrm{O}_{2} at 37^{\circ} \mathrm{C}. The temperature of the mixture is -

Option: 1

32^{\circ} \mathrm{C}


Option: 2

27^{\circ} \mathrm{C}


Option: 3

37^{\circ} \mathrm{C}


Option: 4

30.5^{\circ} \mathrm{C}


Answers (1)

best_answer

Let t be the temperature of the mixture.

Heat gained by $\mathrm{CO}_{2}=Heat lost by \mathrm{O}_{2}

\begin{aligned} & \Rightarrow \mu_{1} c_{v_{1}} \Delta T_{1}=\mu_{2} C_{v_{2}} \Delta T_{2} \\ & \Rightarrow \frac{22}{44}(3 R)(t-27)=\frac{16}{32}\left(\frac{5}{2} R\right)(37-t) \\ & \Rightarrow 3(t-27)=\frac{5}{2}(37-t) \\ & \Rightarrow 3 t-71=\frac{5 \times 37}{2}-\frac{5}{2} t \Rightarrow 3 t+\frac{5}{2} t=\frac{5 \times 37}{2}+71 \end{aligned}

\begin{aligned} & \Rightarrow 3 t+\frac{5}{2} t=\frac{5 \times 37}{2}+71 \\ & \Rightarrow \frac{11}{2} t=\frac{185}{2}+71 \\ & \Rightarrow \frac{11}{2} t=\frac{327}{2} \Rightarrow t=32^{\circ} \mathrm{C} \end{aligned}

 

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