Chapter 12, Problem 78A

To determine

The binding energy per molecule for the substance.

Answer to Problem 78A

The binding energy per molecule for the substance is $4.8\times {10}^{-19}\text{J}/\text{molecule}$ .

Explanation of Solution

Given:

The mass of water is $m=0.50\text{ kg}$ .

The increase in temperature is $\Delta T=2.3°\text{C}$ .

Formula used:

The expression for the amount of energy transferred is

  $\Delta U=mC\Delta T$

Here, $m$ is the mass, $C$ is the specific heat, and $\Delta T$ is the change in temperature.

Calculation:

Consider the specific heat of the water as $C=4,180\text{J}/\text{kg}\cdot \text{°C}$ .

The amount of energy added to the water is,

  $\begin{array}{l}\Delta U=\left(0.50\text{ kg}\right)\times \left(4,180\text{J}/\text{kg}\cdot \text{°C}\right)\times \left(2.3°\text{C}\right)\\ \Delta U=4.8\times {10}^3\text{ J}\end{array}$

  $1.0\times {10}^{22}$ molecules of a powdered substance are dissolved by chemist into water.

The binding energy per molecule for the substance is,

  $\begin{array}{l}\text{Binding energy}=\frac{\Delta U}{\text{Amount of molecules}}\\ \text{Binding energy}=\frac{4.8\times {10}^3\text{ J}}{1.0\times {10}^{22}\text{ molecules}}\\ \text{Binding energy}=4.8\times {10}^{-19}\text{J}/\text{molecule}\end{array}$

Conclusion:

Thus, the binding energy per molecule for the water is $4.8\times {10}^{-19}\text{J}/\text{molecule}$ .