Question

22 grams of $C{O}_2$ at ${27}^{\circ }C$ is mixed with $16g$ of $O_2$ at ${37}^{\circ }C$. The temperature of the mixture is about.

• A. ${31.5}^{\circ }C$
• B. ${27}^{\circ }C$
• C. ${37}^{\circ }C$
• D. ${30.5}^{\circ }C$

Answer 1

The correct option is A ${31.5}^{\circ }C$

Given,

Temperature of $22g$ of $C{O}_2$ is ${27}^{0}C$

Temperature of $16g$ of $O_2$ is ${37}^{0}C$

Let $T$ is the temperature of mixture.

Temperature gain by $C{O}_2$=Temperature loss by $O_2$

${\mu }_1{C}_{v1}{\Delta T}_1={\mu }_2{C}_{v2}{\Delta T}_2$

Substitute all value in above equation

$\frac{22}{44}\times 3R\times (T-27)=\frac{16}{32}\times \left(\frac{5}{2}R\right)(37-T)6R(T-27)=2\left(\frac{5}{2}R\right)(37-T)\frac{6}{5}=\frac{37-T}{T-27}6T-162=185-5T11T=347T={31.5}^{0}C$

Correct option is A.