Chapter 23, Problem 86A

Interpretation Introduction

Interpretation: To determine the amount of heat absorbed by the aluminum tray.

Concept Introduction: Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 gram of a substance by ${\text{1}}^{\text{0}}$ C. When two different materials with different temperatures at the initial stage are kept in contact with each other then there will be a flow of heat. So, when heat is absorbed, there will be a rise in the temperature. If the same quantity of heat is given to equal masses of different substances, then the rise in temperature for each substance will be different.

Answer to Problem 86A

The heat absorbed by the aluminum tray is 71.1 kJ

Explanation of Solution

The heat absorbed by the aluminum tray can be calculated by using the formula

  $\text{Q=mCΔT}$

Where Q is the heat absorbed by the aluminum tray

m is the mass of the aluminum i.e., 500 grams

C is the specific heat of aluminum i.e., 0.9J/(g ${}^{\text{0}}\text{C}$ )

  $\text{ΔT}$ is the change in the temperature

  $\begin{array}{l}\text{Q=500 g}\times \text{0}{\text{.90J/(g}}^{\text{0}}{\text{C)(180}}^{\text{0}}{\text{C-22}}^{\text{0}}\text{C)}\\ \text{Q=500 g }\times \text{0}{\text{.90J/(g}}^{\text{0}}\text{C)}({158}^0\text{C)}\\ \text{Q=71,100J}\\ \text{=71}\text{.1kJ}\end{array}$

Conclusion

1 kJ is 1000 J therefore, the heat absorbed by the 500 gram aluminum tray with 0.9J/(g ${}^{\text{0}}\text{C}$ ) is 71.1kJ.