Chapter 18, Problem 19SP

Victoria Falls on the Zambezi River is 108 m high, and 1088 ${\text{m}}^{\text{3}}$ of water pours over it every second. Assuming no loss in energy, what is the rise in temperature of the water due to the drop? [Hint: Think PE.]

To determine

The temperature rise in water due to drop if 1088 m³ of water is falling from a distance of $108\text{ m}$ every second without any loss in energy.

Answer to Problem 19SP

Solution:

$0.253\text{ K}$

Explanation of Solution

Given data:

Quantity of water falling every second is 1088 m³.

The distance from which the water is falling is $108\text{ m}$.

Formula used:

Write the expression for work potential energy of a substance as

$PE=mgh=\rho Vgh$

Here, $m$ is the mass, $V$ is the volume, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height.

Write the expression for heat gain or lossif a body or substance undergoes temperature change as

$\Delta Q=mc\Delta T$

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

Explanation:

The temperature rise takes place due to the change in potential energy of water.

The expression for potential energy of a substance is as follows:

$PE=\rho Vgh$

The expression for heat required to raise the temperature of water is as follows:

$\Delta Q=mc\Delta T$

From the question, the potential energy is used to raise the temperature of water is

$PE=\Delta Q$

Consider the calculation for unit time:

Substitute $\rho Vgh$ for $W$ and $mc\Delta T$ for $\Delta Q$

$\begin{array}{c}\rho Vgh=\rho Vc\Delta T\\ \Delta T=\frac{gh}{c}\end{array}$

Refer to Table 18-1 in the textbook for the value of specific heat capacity of water as

$c=4.186\text{ kJ/kg}\cdot \text{K}=4186\text{ J/kg}\cdot \text{K}$

Substitute $9.81{\text{ m/s}}^2$ for $g$, $108\text{ m}$ for $h$, and $4186\text{ J/kg}\cdot \text{K}$ for $c$

$\begin{array}{c}\Delta T=\frac{\left(9.81{\text{ m/s}}^2\right)\left(108\text{ m}\right)}{4186\text{ J/kg}\cdot \text{K}}\\ =0.253\text{ K}\end{array}$

Conclusion:

The temperature rise in the waterdue to drop is $0.253\text{ K}$.