Determine the amount of energy Q you need to take off from 320 kg of water if you want to decrease its temperature from 52 °C to 18 °C. (Specific heat of water 4,186 J/Kg K) about 90 J about 45,500,000 J about 90,000,000J about 45,000 J
Determine the amount of energy Q you need to take off from 320 kg of water if you want to decrease its temperature from 52 °C to 18 °C. (Specific heat of water 4,186 J/Kg K) about 90 J about 45,500,000 J about 90,000,000J about 45,000 J
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![**Determine the Energy Needed to Cool Water**
In this exercise, you are tasked with calculating the amount of energy (Q) that must be removed to decrease the temperature of 320 kg of water from 52 °C to 18 °C. The specific heat capacity of water is given as 4,186 J/(kg·K).
**Question:**
Determine the amount of energy Q you need to take off from 320 kg of water if you want to decrease its temperature from 52°C to 18°C. The specific heat of water is 4,186 J/(kg·K).
**Options:**
- About 90 J
- About 45,500,000 J
- About 90,000,000 J
- About 45,000 J
**Explanation:**
To solve for the energy Q, use the formula:
\[ Q = mc\Delta T \]
Where:
- \( Q \) is the energy required (in joules, J)
- \( m \) is the mass of the water (in kilograms, kg)
- \( c \) is the specific heat capacity (in J/(kg·K))
- \( \Delta T \) is the change in temperature (in degrees Celsius, °C or Kelvin, K)
Substitute the given values:
- \( m = 320 \) kg
- \( c = 4,186 \) J/(kg·K)
- \( \Delta T = 52°C - 18°C = 34°C \)
Calculate \( Q \):
\[ Q = 320 \, \text{kg} \times 4,186 \, \frac{\text{J}}{\text{kg·K}} \times 34 \, \text{K} \]
\[ Q = 320 \times 4,186 \times 34 \]
\[ Q = 45,515,520 \, \text{J} \]
Therefore, the correct answer is approximately **45,500,000 J**.
**Correct Answer:**
- About 45,500,000 J](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8025333-2198-4fd3-bdd9-62e65b03c02c%2Fb63603af-1673-4195-b6af-8db07f076270%2Fhks57fg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine the Energy Needed to Cool Water**
In this exercise, you are tasked with calculating the amount of energy (Q) that must be removed to decrease the temperature of 320 kg of water from 52 °C to 18 °C. The specific heat capacity of water is given as 4,186 J/(kg·K).
**Question:**
Determine the amount of energy Q you need to take off from 320 kg of water if you want to decrease its temperature from 52°C to 18°C. The specific heat of water is 4,186 J/(kg·K).
**Options:**
- About 90 J
- About 45,500,000 J
- About 90,000,000 J
- About 45,000 J
**Explanation:**
To solve for the energy Q, use the formula:
\[ Q = mc\Delta T \]
Where:
- \( Q \) is the energy required (in joules, J)
- \( m \) is the mass of the water (in kilograms, kg)
- \( c \) is the specific heat capacity (in J/(kg·K))
- \( \Delta T \) is the change in temperature (in degrees Celsius, °C or Kelvin, K)
Substitute the given values:
- \( m = 320 \) kg
- \( c = 4,186 \) J/(kg·K)
- \( \Delta T = 52°C - 18°C = 34°C \)
Calculate \( Q \):
\[ Q = 320 \, \text{kg} \times 4,186 \, \frac{\text{J}}{\text{kg·K}} \times 34 \, \text{K} \]
\[ Q = 320 \times 4,186 \times 34 \]
\[ Q = 45,515,520 \, \text{J} \]
Therefore, the correct answer is approximately **45,500,000 J**.
**Correct Answer:**
- About 45,500,000 J
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