Question

One litre of water at ${30}^o$C is mixed with one litre of water at ${50}^o$C. The temperature of the mixture will be.

• A. ${80}^o$C
• B. More than ${50}^o$C but less than ${80}^o$C
• C. ${20}^o$C
• D. Between ${30}^o$C and ${50}^o$C

Answer 1

The correct option is (D). Between ${30}^o$C and ${50}^o$C.

Let the first volume of water be $V_1$. Let the second volume of water be $V_2$. Let the initial temperature of the first volume of water be ${\theta }_1$. Let the initial temperature of the second volume of water be ${\theta }_2$

Let the energy required to heat a unit volume of water by unit temperature be $C$.

So, if the final temperature is ${\theta }_{f}inal$, we have heat gained by the first volume of water =$V_1\times C\times ({\theta }_{f}inal-{\theta }_1)$

and heat lost by the second volume of water =$V_2\times C\times ({\theta }_2-{\theta }_{f}inal)$

By the law of conservation of energy,

heat lost = heat gained

∴$V_1\times C\times ({\theta }_{f}inal-{\theta }_1)=V_2\times C\times ({\theta }_2-{\theta }_{f}inal)$

∴$V_1{\theta }_{f}inal-V_1{\theta }_1=V_2{\theta }_2-V_2{\theta }_{f}inal$

∴${\theta }_{f}inal=\frac{V_1{\theta }_1+V_2{\theta }_2}{V_1+V_2}$

Substituting the given values,

${\theta }_{f}inal=\frac{30+50}{2}={40}^{\circ }C$.

The answer is option (D).