Chapter 10, Problem 55A

Interpretation Introduction

Interpretation: The final temperature of water-iron mixture needs to be determined, if 25.0 g of pure iron at 85°C is dropped in to 75 g of water at 20°C.

Concept Introduction: Specific heat capacity can be defined as the heat in joules that is required to raise the temperature of 1 g of substance by 1°C. The relation between specific heat capacity and heat required is shown below:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = heat }\\ \text{m = mass }\\ \text{C = specific heat capacity }\\ \text{ΔT= temperature change}\end{array}$

Answer to Problem 55A

  ${\text{T}}_{\text{f}}\text{= 17}\text{.6°C}$

Explanation of Solution

Given information:

Mass of pure iron = 25.0 g

Initial temperature of pure iron = 85°C

Mass of water = 75 g

Initial temperature of water = 20°C

Specific heat capacity of iron = 0.45 J/g°C

Specific heat capacity of water = 4.184 J/g°C

The heat lose by the metal must be heat gain by water. Thus the equation must be:

  $\begin{array}{l}{\text{q}}_{\text{water}}{\text{ = q}}_{\text{iron}}\\ {\text{(m× C × ΔT)}}_{\text{water}}{\text{ = (m× C × ΔT)}}_{\text{iron}}\end{array}$

Substitute the values to calculate the final temperature:

  $\begin{array}{l}{\text{q}}_{\text{water}}{\text{ = q}}_{\text{iron}}\\ \text{(75 g × 4}{\text{.184 J/g°C × (T}}_{\text{f}}{\text{ - 20°C)}}_{\text{water}}\text{ = (25}\text{.0 g× 0}{\text{.45J/g°C × (T}}_{\text{f}}{\text{ - 85°C)}}_{\text{iron}}\\ $ {\text{(313}{\text{.8T}}_{\text{f}}\text{-6276)}}_{\text{water}}\text{ = (11}{\text{.25T}}_{\text{f}}\text{-956}{\text{.25)}}_{\text{iron}}$\text{313}{\text{.8T}}_{\text{f}}\text{-11}{\text{.25T}}_{\text{f}}\text{=6276-956}\text{.25}\\ \text{302}{\text{.55T}}_{\text{f}}\text{=5319}\text{.75}\\ {\text{T}}_{\text{f}}\text{=}\frac{\text{5319}\text{.75}}{\text{302}\text{.55}}\text{°C}\\ {\text{T}}_{\text{f}}\text{= 17}\text{.6°C}\end{array}$

Conclusion

Thus, the value of ${\text{T}}_{\text{f}}\text{= 17}\text{.6°C}$