Chapter 10, Problem 9A

Interpretation Introduction

Interpretation: The case in which more heat is involved needs to be determined.

Concept Introduction:

When two solutions or water of different temperature in mixed, the warm solution will become cold and the cold solution will become warm or both of them will come at a same temperature.

Answer to Problem 9A

Mixing 100.0 g sample of 60-degree Celsius water and 10-degree Celsius water.

Explanation of Solution

The energy which flowed out is always equals to the energy which is flowed in.

Now,

Lets us assume that the final temperature is C.

Therefore,

Total energy lost = total energy gained

Also,

Total energy lost when the sample is 100g $=100\left(80-C\right)$

Total energy gained when the sample is 60g $=\text{ 60}\left(C-20\right)$

So, putting the value in the above equation,

  $\begin{array}{l}100(80-C)=60(C-20)\\ 8000-100C=60C-1200\\ 8000+1200=60C+100C\\ 9200=160C\\ C=\frac{9200}{160}\\ C=57.5\simeq 60\end{array}$

Mixing 100.0 g sample of 60-degree Celsius water and 10-degree Celsius water.

Conclusion

Thus, mixing 100.0_g sample of 60-degree Celsius water and 10-degree Celsius water will involve more heat.