QuestionA power station gets rid of 600 MW of surplus energy into a nearby river. The river is 20 m wide, 3.0 m deep and flows at $1.2 \mathrm{~m} \mathrm{~s}^{-1}$. Calculate the temperature rise.

Studdy Solution
Use the energy balance to find the temperature rise.
The energy added to the river is $600 \, \mathrm{MW} = 600 \times 10^6 \, \mathrm{J/s}$.
The temperature rise $\Delta T$ can be calculated using the formula:
$$P = \dot{m} \times c \times \Delta T$$
Where $P$ is the power added, $\dot{m}$ is the mass flow rate, $c$ is the specific heat capacity of water (approximately $4186 \, \mathrm{J/(kg \cdot °C)}$), and $\Delta T$ is the temperature rise.
Rearranging for $\Delta T$:
$$\Delta T = \frac{P}{\dot{m} \times c} = \frac{600 \times 10^6 \, \mathrm{J/s}}{72000 \, \mathrm{kg/s} \times 4186 \, \mathrm{J/(kg \cdot °C)}}$$
$$\Delta T \approx \frac{600 \times 10^6}{72000 \times 4186}$$
$$\Delta T \approx 1.99 \, °C$$
The temperature rise of the river water is approximately:
$\boxed{1.99 \, °C}$