Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let t is the temperature of mixture
Heat gained by$C{O}_2$ = Heat lost by $O_2$
$\Rightarrow {\mu }_1{C_v}_1\Delta T_1={\mu }_2{C_v}_2\Delta T_2$
$\Rightarrow \frac{22}{44}\left(3R\right)(t-27)=\frac{16}{32}\left(\frac{5}{2}R\right)(37-t)$
$\Rightarrow 3(t-27)=\frac{5}{2}(37-t)$
By solving we get $t={32}^{\circ }C.$