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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**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
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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