Chapter 10, Problem 10A

Interpretation Introduction

Interpretation :

Final temperature of the water must be calculated when 100 g water at 100°C is mixed with 50 g water at 10°C.

Concept Introduction :

According to Calorimetry principle, hot water releases heat which is absorbed by the cold water. This heat transfer process continues until the temperature of the two water sample becomes same.

Final temperature $\text{=}\frac{{\text{m}}_{\text{1}}{\text{×t}}_{\text{1}}{\text{+m}}_{\text{2}}{\text{×t}}_{\text{2}}}{{\text{m}}_{\text{1}}{\text{+m}}_{\text{2}}}\text{.}$

Here m₁, m₂ are masses of two water samples and t₁, t₂ are temperatures of the two water samples respectively.

Answer to Problem 10A

Final temperature of the mixture of water is 63.33°C.

Explanation of Solution

Masses of two water samples are 100 g and 50 g respectively.

Temperatures of the two water samples are 90C and 50C respectively.

Putting all the values,

Final temperature $\text{=}\frac{\text{100×90+50×10}}{\text{100+50}}°\text{C=}\frac{\text{9000+500}}{\text{150}}°C=\frac{9500}{150}°C=63.33°C.$

Conclusion

Final temperature of the water mixture is 63.33°C.