Q. 1.41. - An Introduction to Thermal Physics

Question

Problem 1.41. To measure the heat capacity of an object, all you usually have to do is put it in thermal contact with another object whose heat capacity you know. As an example, suppose that a chunk of metal is immersed in boiling water (100°C), then is quickly transferred into a Styrofoam cup containing 250 g of water at 20°C. After a minute or so, the temperature of the contents of the cup is 24°C. Assume that during this time no significant energy is transferred between the contents of the cup and the surroundings. The heat capacity of the cup itself is negligible.

1. How much heat is lost by the water?
2. How much heat is gained by the metal?
3. What is the heat capacity of this chunk of metal?
4. If the mass of the chunk of metal is 100 g, what is its specific heat capacity?

Step-by-Step Solution

Answer

1. The heat gained by water is 1000 cal.
2. -1000 cal.
3. $13.157{\text{\hspace{0.17em}cal.K}}^{\text{-1}}$
   
4. The specific heat of 100 g of a chunk of the metal is $c=0.131{\text{cal.g}}^{\text{-1}}{\text{K}}^{\text{-1}}$.

Part a. Step 1. Given.

Mass of water, $m=250\text{g}$

Specific heat of water, $c=1{\text{\hspace{0.17em}cal.g}}^{\text{-1}}{\text{K}}^{\text{-1}}$

Initial temperature, $T_1={24}^0\text{C}$

Final temperature, $T_2={20}^0\text{C}$

Part a. Step 2. Calculation.

The first energy transfer takes place between the chunk of metal and the boiling water at temperature ${100}^0\text{C}$. The next energy transfer takes place between the metal and the water in the Styrofoam cup that was at temperature ${20}^0\text{C}$.

The heat absorbed $Q$ by the water from the hot metal is given by the expression,

$\begin{array}{l}Q=mc\left(T_2-T_1\right)\\ Q=250\times 1\times \left(4\right)\\ \boxed{Q=1000\text{\hspace{0.17em}cal}}\end{array}$

Part a. Step 3. Conclusion.

The only energy transfer is heat transfer, thus the heat absorbed by the water is 1000 cal.

Part b. Step 1. Introduction.

Heat capacity – It is defined as the ratio of heat absorbed to the change in the temperature.

Specific heat – It is the heat that is needed to raise the temperature of a unit mass to $1^0\text{C}$. 

Energy transfer between two bodies is due to the difference in temperature of the bodies.   

Part b. Step 2. Explanation.

The heat transfer takes place until the two bodies acquire an equilibrium state and the amount of heat absorbed by water is equal to the heat lost by the metal.

Part b. Step 3. Conclusion.

The heat lost by the metal is the heat absorbed by the water, that is -1000 cal.

Part c. Step 1. Given.

Initial temperature, $T_1={100}^{\text{0}}\text{C}$

Final temperature, $T_2={24}^{\text{0}}\text{C}$

Amount of energy gained by the chunk, $\text{Q}=-1000\text{cal}$

Part c. Step 2. Calculation.

The heat capacity of the metal is given by the expression,

$\begin{array}{l}C=\frac{Q}{T_2-T_1}\\ C=\frac{-1000}{\begin{array}{l}24-100\\ =\frac{-1000}{-76}\\ =13.157\text{cal}\cdot {\text{K}}^{\text{-1}}\end{array}}\end{array}$

Part c. Step 3. Conclusion.

The heat capacity of the metal chunk is calculated as $\boxed{C=13.157{\text{cal.g}}^{\text{-1}}{\text{K}}^{\text{-1}}}$.

Part d. Step 1. Given.

Heat capacity of the metal, $C=13.157\text{cal}\cdot {\text{K}}^{\text{-1}}$

Mass of the metal, $m=100\text{\hspace{0.17em}g}$

Part d. Step 2. Calculation.

$\begin{array}{l}c=\frac{C}{m}\\ c=\frac{13.157}{100}\\ c=0.131{\text{cal.g}}^{\text{-1}}{\text{K}}^{\text{-1}}\end{array}$

Part d. Step 3. Conclusion.

Specific heat capacity of the 100 g mass of metal is calculated as $\boxed{c=0.131{\text{cal.g}}^{\text{-1}}{\text{K}}^{\text{-1}}}$.