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.)
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.)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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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.
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This snippet would help students understand and calculate the net heat transfer during the heating of a swimming pool using fundamental thermodynamic equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2F1d007120-203b-43de-8549-b6d45c580edf%2Fceapyk.png&w=3840&q=75)
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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