Chapter 12.2, Problem 35SSC

To determine

To Calculate: The temperature difference of water at top and bottom of the waterfall.

Answer to Problem 35SSC

  $0.293\text{°C}$

Explanation of Solution

Given:

The height of the waterfall, $h=125.0\text{m}$

Formula Used:

Heat required to raise the temperature can be obtained by:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

Also, work done when the bag is dropped = gravitational potential energy.

  $W=mgh$

Where, m is the mass, g is the gravitational acceleration and h is the height.

Calculations:

Specific heat capacity of water is $c=4184\text{J/kg}\cdot \text{°C}$

Amount of heat required to raise the temperature of water is:

  $\begin{array}{c}Q=mc\Delta T\\ =m\times 4184\text{J/kg}\cdot \text{°C}\times \Delta T\end{array}$

Amount of heat generated when water falls:

  $\begin{array}{c}W=mgh\\ =m\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\left(\text{125}\text{m}\right)\end{array}$

  $\begin{array}{l}Q=W\\ m\left(4184\right)\left(\Delta T\right)=m\left(9.8\right)\left(125\right)\\ \Delta T=\frac{\left(9.8\right)\left(125\right)}{\left(4184\right)}\\ \Delta T=0.293\text{°C}\end{array}$

Conclusion: The temperature difference of water at top and bottom of the waterfall is $0.293\text{°C}$ .