On a hot day, the temperature of a 76,000 L swimming pool increases by 1.10°C. What is the net heat transfer (in J) during this heating? Ignore any complications, such as loss of water by evaporation. (Enter the magnitude.)

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### Determining the Net Heat Transfer in a Swimming Pool

On a hot day, the temperature of a 76,000 L swimming pool increases by 1.10°C. What is the net heat transfer (in J) during this heating? Ignore any complications, such as loss of water by evaporation. (Enter the magnitude.)

\[ \boxed{} \text{ J} \]

To solve this problem, we need to use the formula for heat transfer:
\[ Q = mc\Delta T \]
where:
- \( Q \) is the heat energy in joules (J),
- \( m \) is the mass of the water in kilograms (kg),
- \( c \) is the specific heat capacity of water (which is approximately 4,186 J/kg°C),
- \( \Delta T \) is the change in temperature in degrees Celsius (°C).

Given the volume of water and the temperature change:
- Volume \( V \) = 76,000 L (Since 1 L of water has a mass of approximately 1 kg, the mass \( m \) is 76,000 kg),
- Temperature change \( \Delta T \) = 1.10°C.

We can now plug the values into the formula:
\[ Q = 76,000 \, \text{kg} \times 4,186 \, \text{J/kg°C} \times 1.10°C \]

Simplify to find \( Q \):
\[ Q = 76,000 \times 4,186 \times 1.10 \]

The resulting value of \( Q \) is the net heat transfer in joules. Enter the calculated magnitude in the provided box.

---

This snippet would help students understand and calculate the net heat transfer during the heating of a swimming pool using fundamental thermodynamic equations.
Transcribed Image Text:### Determining the Net Heat Transfer in a Swimming Pool On a hot day, the temperature of a 76,000 L swimming pool increases by 1.10°C. What is the net heat transfer (in J) during this heating? Ignore any complications, such as loss of water by evaporation. (Enter the magnitude.) \[ \boxed{} \text{ J} \] To solve this problem, we need to use the formula for heat transfer: \[ Q = mc\Delta T \] where: - \( Q \) is the heat energy in joules (J), - \( m \) is the mass of the water in kilograms (kg), - \( c \) is the specific heat capacity of water (which is approximately 4,186 J/kg°C), - \( \Delta T \) is the change in temperature in degrees Celsius (°C). Given the volume of water and the temperature change: - Volume \( V \) = 76,000 L (Since 1 L of water has a mass of approximately 1 kg, the mass \( m \) is 76,000 kg), - Temperature change \( \Delta T \) = 1.10°C. We can now plug the values into the formula: \[ Q = 76,000 \, \text{kg} \times 4,186 \, \text{J/kg°C} \times 1.10°C \] Simplify to find \( Q \): \[ Q = 76,000 \times 4,186 \times 1.10 \] The resulting value of \( Q \) is the net heat transfer in joules. Enter the calculated magnitude in the provided box. --- This snippet would help students understand and calculate the net heat transfer during the heating of a swimming pool using fundamental thermodynamic equations.
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