Chapter U5.97, Problem 6E

Interpretation Introduction

Interpretation: The final temperature of the water of the mixer of water sample which is formed by mixing 300 g of water 20°C and 200 g of water at 50°C needs to be determined.

Concept Introduction: Due to transfer of energy the temperature of any substance increases or decreases. Specific heat can be defined as the heat that would require to raise the temperature of 1 unit mass by per °C. The change in temperature for the addition of energy can be calculated with the help of calorimetric equation:

  $\begin{array}{l}\text{Q = m × C × ΔT}\\ \text{Here:}\\ \text{Q}\text{}\text{ = heat change }\\ \text{m = mass}\\ \text{C = specific heat }\\ \text{ΔT=change in T}\end{array}$

Answer to Problem 6E

  ${\text{T}}_{\text{3}}\text{=32°C}$

Explanation of Solution

Mass of water of sample 1= 300 g

Temperature of water of sample 1 = 20°C

Mass of water of sample 1= 200 g

Temperature of water of sample 1 = 50°C

Total mass of mixture = 200 + 300 = 500 g

  $\begin{array}{l}{\text{(m}}_{\text{1}}{\text{×T}}_{\text{1}}{\text{)+(m}}_{\text{2}}{\text{×T}}_{\text{2}}{\text{)=(m}}_{\text{3}}{\text{×T}}_{\text{3}}\text{)}\\ {\text{(300 g×20°C)+(200 g×50°C)=(500 g×T}}_{\text{3}}\text{)}\\ {\text{ T}}_{\text{3}}\text{=}\frac{\text{(300 g×20°C)+(200 g×50°C)}}{\text{500 g}}\\ {\text{ T}}_{\text{3}}\text{=32°C}\end{array}$

Conclusion

Thus, the final temperature of mixture must be 32°C.