Chapter 21, Problem 28A

When a 50-gram piece of aluminum at $100°C$ is placed in water, it loses 735 calories of heat while cooling to $30°C$. Calculate the specific heat capacity of the aluminum.

To determine

Specific heat capacity of aluminum.

Answer to Problem 28A

Specific heat capacity of aluminum is, $C=0.214\text{cal/g}°\text{C}$ .

Explanation of Solution

Given:

Mass of aluminum, $m=50\text{g}$

Initial temperature of aluminum is $100°\text{C}$

Final temperature of aluminum is $30°\text{C}$

The number of calories given loses by aluminum, $Q=735\text{cal}$

Formula Used:

The heat transfer in calories is given by,

  $Q=mC\Delta T$

Where, $m$ is mass in grams, $C$ is specific heat capacity in $\text{cal/g}°\text{C}$ and $\Delta T$ is change in temperature in $°\text{C}$ .

Calculation:

The change in temperature is,

  $\Delta T=100°\text{C}-30°\text{C}=70°\text{C}$

Now, putting the values in heat transfer equation

  $\begin{array}{l}Q=mC\Delta T\\ 735=\left(50\text{g}\right)\times C\times \left(70°\text{C}\right)\\ \Rightarrow C=\frac{750}{50\times 70}\text{g}\\ \Rightarrow C=0.214\text{cal/g}°\text{C}\end{array}$

Conclusion:

Therefore, the specific heat capacity of aluminum is, $C=0.214\text{cal/g}°\text{C}$ .