Question
The total amount of energy released when 45 g of 175 ° C' steam is cooled to 90°C is _ Reminder: As steam is condensed it will turn into water at 100 ° C. Round your answer to the nearest whole number. (1 point) A. 417,600 J B. 317,781 J C.110,432 J D.101,700 J
Which is the correct answer?
Which is the correct answer?
Answers
GPT-4o mini
To calculate the total amount of energy released when 45 g of steam at 175 °C is cooled to 90 °C, let's break it down into the various phases of heat exchange:
1. **Cooling the steam from 175 °C to 100 °C**
2. **Condensing the steam at 100 °C to water**
3. **Cooling the water from 100 °C to 90 °C**
**Given Data:**
- Mass of steam (m) = 45 g = 0.045 kg
- Specific heat of steam (c_steam) = 2,009 J/(kg·°C)
- Specific heat of water (c_water) = 4,186 J/(kg·°C)
- Heat of vaporization (L_vaporization) = 2,260,000 J/kg
### Step 1: Cooling the steam from 175 °C to 100 °C
\[
q_1 = m \cdot c_{\text{steam}} \cdot \Delta T
\]
\[
q_1 = 0.045 \, \text{kg} \cdot 2,009 \, \text{J/(kg·°C)} \cdot (100 - 175) \, °C
\]
\[
q_1 = 0.045 \cdot 2,009 \cdot (-75)
\]
\[
q_1 = -6,773.25 \, \text{J}
\]
### Step 2: Condensing the steam at 100 °C
\[
q_2 = m \cdot L_{\text{vaporization}}
\]
\[
q_2 = 0.045 \, \text{kg} \cdot 2,260,000 \, \text{J/kg}
\]
\[
q_2 = 101,700 \, \text{J}
\]
### Step 3: Cooling the water from 100 °C to 90 °C
\[
q_3 = m \cdot c_{\text{water}} \cdot \Delta T
\]
\[
q_3 = 0.045 \, \text{kg} \cdot 4,186 \, \text{J/(kg·°C)} \cdot (90 - 100) \, °C
\]
\[
q_3 = 0.045 \cdot 4,186 \cdot -10
\]
\[
q_3 = -1,883.7 \, \text{J}
\]
### Total Energy Released
Sum up all the heat changes (q values):
\[
q_{\text{total}} = q_1 + q_2 + q_3
\]
- Since \(q_1\) and \(q_3\) are negative (heat lost) and \(q_2\) is positive (heat lost due to condensation):
\[
q_{\text{total}} = -6,773.25 + 101,700 - 1,883.7
\]
\[
q_{\text{total}} = 93,043.05 \, \text{J}
\]
### Rounding Off
Rounding the total energy released to the nearest whole number gives:
\[
q_{\text{total}} \approx 93,043\, \text{J}
\]
### Conclusion
None of the provided options match the computation explicitly. However, the closest option would usually be checked against rounding allowances or calculation errors within standard estimations.
However, based on the options you provided, it seems \(D.\, 101,700\, J\) corresponds specifically to the condensation phase only. The full calculation for the total heat released shows a larger value.
Please double-check the options or clarify if any phases or specific calculations may be omitted. If you're looking strictly for total energy released based on full condensation into liquid at 100 °C and then cooling to 90 °C without the interim cooling, \(D\) stands correct for the condensation phase itself.
1. **Cooling the steam from 175 °C to 100 °C**
2. **Condensing the steam at 100 °C to water**
3. **Cooling the water from 100 °C to 90 °C**
**Given Data:**
- Mass of steam (m) = 45 g = 0.045 kg
- Specific heat of steam (c_steam) = 2,009 J/(kg·°C)
- Specific heat of water (c_water) = 4,186 J/(kg·°C)
- Heat of vaporization (L_vaporization) = 2,260,000 J/kg
### Step 1: Cooling the steam from 175 °C to 100 °C
\[
q_1 = m \cdot c_{\text{steam}} \cdot \Delta T
\]
\[
q_1 = 0.045 \, \text{kg} \cdot 2,009 \, \text{J/(kg·°C)} \cdot (100 - 175) \, °C
\]
\[
q_1 = 0.045 \cdot 2,009 \cdot (-75)
\]
\[
q_1 = -6,773.25 \, \text{J}
\]
### Step 2: Condensing the steam at 100 °C
\[
q_2 = m \cdot L_{\text{vaporization}}
\]
\[
q_2 = 0.045 \, \text{kg} \cdot 2,260,000 \, \text{J/kg}
\]
\[
q_2 = 101,700 \, \text{J}
\]
### Step 3: Cooling the water from 100 °C to 90 °C
\[
q_3 = m \cdot c_{\text{water}} \cdot \Delta T
\]
\[
q_3 = 0.045 \, \text{kg} \cdot 4,186 \, \text{J/(kg·°C)} \cdot (90 - 100) \, °C
\]
\[
q_3 = 0.045 \cdot 4,186 \cdot -10
\]
\[
q_3 = -1,883.7 \, \text{J}
\]
### Total Energy Released
Sum up all the heat changes (q values):
\[
q_{\text{total}} = q_1 + q_2 + q_3
\]
- Since \(q_1\) and \(q_3\) are negative (heat lost) and \(q_2\) is positive (heat lost due to condensation):
\[
q_{\text{total}} = -6,773.25 + 101,700 - 1,883.7
\]
\[
q_{\text{total}} = 93,043.05 \, \text{J}
\]
### Rounding Off
Rounding the total energy released to the nearest whole number gives:
\[
q_{\text{total}} \approx 93,043\, \text{J}
\]
### Conclusion
None of the provided options match the computation explicitly. However, the closest option would usually be checked against rounding allowances or calculation errors within standard estimations.
However, based on the options you provided, it seems \(D.\, 101,700\, J\) corresponds specifically to the condensation phase only. The full calculation for the total heat released shows a larger value.
Please double-check the options or clarify if any phases or specific calculations may be omitted. If you're looking strictly for total energy released based on full condensation into liquid at 100 °C and then cooling to 90 °C without the interim cooling, \(D\) stands correct for the condensation phase itself.