Question

22 g 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

• A. ${27}^{\circ }C$
• B. ${30.5}^{\circ }C$
• C. ${32}^{\circ }C$
• D. ${37}^{\circ }C$

Answer 1

The correct option is C ${32}^{\circ }C$

Heat lost by $C{O}_2$ = Heat gained by $0_2$
$If{\mu }_{1}and{\mu }_2$ are the number of moles of carbon di-oxide and oxygen respectively and $C_{v1}and{C}_{v2}$ are the specific heats at constant volume then ${\mu }_1{C}_{v1}\Delta T_1={\mu }_2{C}_{v2}\Delta T_2$
$\Rightarrow \frac{22}{44}\times 3R\times (T-27)=\frac{16}{32}\times \frac{5R}{2}(37-T)\Rightarrow T={31.5}^{\circ }C\approx {32}^{\circ }C$
(where T is temperature of mixture)