Part B Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 48 MW of electricity? This is a typical value for a small hydroelectric dam,

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### Educational Content: Energy Conversion in Hydroelectric Dams

#### Context
In a hydroelectric dam, water falls 35 meters and then spins a turbine to generate electricity.

#### Part A

**Question:** What is the change in potential energy (\(\Delta U\)) of 1.0 kg of water?

*Express your answer with the appropriate units.*

**Solution:**
The change in potential energy of 1.0 kg of water in falling 35 m is calculated as:
\[ \Delta U = -mgh = -(1.0 \, \text{kg})(9.8 \, \text{m/s}^2)(35 \, \text{m}) = -340 \, \text{J} \]

- **Result:** \(\Delta U = -340 \, \text{J}\)

(The answer is marked correct.)

#### Part B

**Problem Statement:**
Suppose the dam is 80% efficient at converting the water’s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 48 MW of electricity? This is a typical value for a small hydroelectric dam.

*Express your answer in kilograms per second.*

**Input Section:**
An equation box is provided where the student is expected to input the required mass flow rate in \(\, \text{kg/s}\).

**Provide Feedback:**
Links are available for submitting the answer and accessing hints or feedback.

---

### Additional Explanation:
In Part A, the concept of gravitational potential energy is explored. The formula \(\Delta U = -mgh\) is used to find the energy change due to the fall of water.

Part B focuses on energy conversion efficiency. The task requires understanding the relationship between potential energy, efficiency, and power output to find water mass flow rate.
Transcribed Image Text:### Educational Content: Energy Conversion in Hydroelectric Dams #### Context In a hydroelectric dam, water falls 35 meters and then spins a turbine to generate electricity. #### Part A **Question:** What is the change in potential energy (\(\Delta U\)) of 1.0 kg of water? *Express your answer with the appropriate units.* **Solution:** The change in potential energy of 1.0 kg of water in falling 35 m is calculated as: \[ \Delta U = -mgh = -(1.0 \, \text{kg})(9.8 \, \text{m/s}^2)(35 \, \text{m}) = -340 \, \text{J} \] - **Result:** \(\Delta U = -340 \, \text{J}\) (The answer is marked correct.) #### Part B **Problem Statement:** Suppose the dam is 80% efficient at converting the water’s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 48 MW of electricity? This is a typical value for a small hydroelectric dam. *Express your answer in kilograms per second.* **Input Section:** An equation box is provided where the student is expected to input the required mass flow rate in \(\, \text{kg/s}\). **Provide Feedback:** Links are available for submitting the answer and accessing hints or feedback. --- ### Additional Explanation: In Part A, the concept of gravitational potential energy is explored. The formula \(\Delta U = -mgh\) is used to find the energy change due to the fall of water. Part B focuses on energy conversion efficiency. The task requires understanding the relationship between potential energy, efficiency, and power output to find water mass flow rate.
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