Chapter 10, Problem 56A

Interpretation Introduction

Interpretation: The final temperature of the resultant water-iron mixture needs to be determined if a sample of $\text{25}\text{.0}\text{g}$ pure iron at $85.0°\text{C}$ is dropped into $\text{75}\text{.0}\text{g}$ of water sample at $20.0°\text{C}$ .

Concept introduction: Specific heat of capacity is the amount of heat required to raise the temperature of a given substance of mass, m 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 56A

The final temperature of the resultant water-iron mixture will be $22\text{°C}$ .

Explanation of Solution

Initially, the temperature of iron sample is $20.0°\text{C}$ and that of water sample is $85.0°\text{C}$ . Mass of pure iron and water samples are $\text{25}\text{.0}\text{g}$ and $\text{75}\text{.0}\text{g}$ respectively.

The specific heat capacity for the liquid water is $4.184\text{J/g°C}$ and that of iron is $0.45\text{J/g°C}$ .

Let the final temperature of the resultant water-iron mixture be $T_2$ .

The heat lost by the iron sample, which is at $85.0°\text{C}$ can be determined by using equation (1) as:

  $Q_{lost}=0.45\text{J/g°C}\times 25.0\text{g}\times \left(T_2-85.0\right)\text{°C}$

The heat gained by water sample, which is at $20.0°\text{C}$ can be determined by using equation (1) as:

  $Q_{gain}=4.184\text{J/g°C}\times 75.0\text{g}\times \left(T_2-20.0\right)\text{°C}$

The heat lost by the iron sample will be equal to the amount of heat gained by water sample as:

  $Q_{gain}=-Q_{lost}$

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

  $\begin{array}{c}4.184\text{J/g°C}\times 75.0\text{g}\times \left(T_2-20.0\right)\text{°C}=-0.45\text{J/g°C}\times 25.0\text{g}\times \left(T_2-85.0\right)\text{°C}\\ 27.9\left(T_2-20.0\right)\text{°C}=-\left(T_2-85.0\right)\text{°C}\\ 28.9\times T_2=643\text{°C}\\ T_2=22\text{°C}\end{array}$

Hence, the final temperature of the resultant water-iron mixture will be $22\text{°C}$ .

Conclusion

After mixing the both water samples, the final temperature of the resultant mixture of water-iron will be in between the two given temperatures (i.e. between $20.0°\text{C}$ and $85.0°\text{C}$ ) which means one sample will lose heat and another sample will gain heat. Heat lost by the iron sample will be equal to the amount of heat gained by the water sample.