Chapter 9, Problem 1TS

What would be the final temperature of a mixture of 50 gram $20\text{°C}$ of water and 50 g of $40\text{°C}$ water?

Solution

To determine:The final temperature of a mixture of 50 gram of $20\text{°C}$ water and 50 g of $40\text{°C}$ water

Answer:

The final temperature of a mixture of 50 gram of $20\text{°C}$ water and 50 g of $40\text{°C}$ water is $30\text{°C}$ .

Explanation:

Given info: Mass of the water at $20\text{°C}$ is 50 gram and the mass of the water at $\text{40°C}$ is 50 g .

Write the expression for the heat transfer.

$m_{1}s\left(T-T_1\right)=m_{2}s\left(T_2-T\right)$

Here,

m₁ is the mass of the water at $20\text{°C}$

m₂ is the mass of the water at $40\text{°C}$

s is the specific heat capacity of the water

T₁ is the temperature of the water before mixing

T₂ is the temperature of the water before mixing

T is the final temperature of the mixture

Substitute 50 g for m₁, 50 g for m₂ ,$20\text{°C}$ for T₁ and $40\text{°C}$ for T₂ in the above expression to get T .

$\begin{array}{c}50\text{g}\times s\left(T-20\text{°C}\right)=50\text{g}\times s\left(40\text{°C}-T\right)\\ 2T=60°\text{C}\\ T=30°\text{C}\end{array}$

Conclusion:

Therefore, the final temperature of a mixture of 50 gram of $20\text{°C}$ water and 50 g of $40\text{°C}$ water is $30\text{°C}$ .