Chapter 10, Problem 55A

Interpretation Introduction

Interpretation: The final temperature of water needs to be determined if a sample of $\text{50}\text{.0}\text{g}$ water at $100.0°\text{C}$ is poured into another sample of $\text{50}\text{.0}\text{g}$ water at $25.0°\text{C}$ .

Concept introduction: Specific heat of capacity can be defined as the amount of heat added to a substance to raise the temperature by one degree.

The equation for heat is as follows:

  $Q=s\times m\times \Delta T\mathrm{......}\left(1\right)$

Where, Q is the amount of heat, m is the mass of substance, s is the specific heat capacity and $\Delta T$ is the temperature change.

Answer to Problem 55A

The final temperature of water will be $62\text{°C}$ .

Explanation of Solution

Initially, the temperature of both the water samples are $100.0°\text{C}$ and $25.0°\text{C}$ . Mass of each water sample is $\text{50}\text{.0}\text{g}$ .

The specific heat capacity for the liquid water is $4.184\text{J/g°C}$ .

Let the final temperature be $T_2$ .

The heat lost by the water sample, which is at $100.0°\text{C}$ can be determined by using equation (1) as:

  $Q_{lost}=4.184\text{J/g°C}\times 50.0\text{g}\times \left(T_2-100.0\right)\text{°C}$

The heat gained by the water sample, which is at $25.0°\text{C}$ can be determined by using equation (1) as:

  $Q_{gain}=4.184\text{J/g°C}\times 50.0\text{g}\times \left(T_2-25.0\right)\text{°C}$

The heat lost by the one water sample will be equal to the heat gained by the another water sample as:

  $Q_{gain}=-Q_{lost}$

Where, the negative sign represents that the heat is released during the process.

  $\begin{array}{c}4.184\text{J/g°C}\times 50.0\text{g}\times \left(T_2-25.0\right)\text{°C}=4.184\text{J/g°C}\times 50.0\text{g}\times \left(T_2-100.0\right)\text{°C}\\ \left(T_2-25.0\right)\text{°C}=-\left(T_2-100.0\right)\text{°C}\\ 2T_2=125\text{°C}\\ T_2=62\text{°C}\end{array}$

Hence, the final temperature of the resultant mixture will be $62\text{°C}$ .

Conclusion

After mixing the both water samples, the final temperature of the resultant mixture will be in between the two given temperatures (i.e. between $100.0°\text{C}$ and $25.0°\text{C}$ ) which means one sample will lose heat and another sample will gain heat. Heat gained by the one water sample will be equal to the heat lost by the another water sample.