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
Is the correct answer D?

Answer 1

To calculate the total energy released when 45 g of steam at 175 °C is cooled to 90 °C, we need to consider several steps: cooling the steam to 100 °C, condensing it to water at 100 °C, and then cooling the water from 100 °C to 90 °C.

1. **Cooling the steam from 175 °C to 100 °C:**
\[
q_1 = m \cdot c_{\text{steam}} \cdot \Delta T
\]
Where:
- \( m = 45 \, \text{g} \)
- \( c_{\text{steam}} = 2.0 \, \text{J/g°C} \) (specific heat of steam)
- \( \Delta T = 100 - 175 = -75 \, \text{°C} \)

\[
q_1 = 45 \, \text{g} \cdot 2.0 \, \text{J/g°C} \cdot (-75 \, \text{°C}) = -6750 \, \text{J}
\]

2. **Condensing the steam at 100 °C into water at 100 °C:**
\[
q_2 = m \cdot L_{\text{vaporization}}
\]
Where:
- \( L_{\text{vaporization}} = 2260 \, \text{J/g} \) (latent heat of vaporization)

\[
q_2 = 45 \, \text{g} \cdot 2260 \, \text{J/g} = 101700 \, \text{J}
\]

Note that this value is positive because we are calculating energy released from the steam as it condenses.

3. **Cooling the water from 100 °C to 90 °C:**
\[
q_3 = m \cdot c_{\text{water}} \cdot \Delta T
\]
Where:
- \( c_{\text{water}} = 4.18 \, \text{J/g°C} \) (specific heat of water)
- \( \Delta T = 90 - 100 = -10 \, \text{°C} \)

\[
q_3 = 45 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (-10 \, \text{°C}) = -1881 \, \text{J}
\]

4. **Total energy released:**
The total energy released is given by:
\[
Q_{\text{total}} = q_1 + q_2 + q_3
\]
Substituting the values we calculated:
\[
Q_{\text{total}} = -6750 + (-101700) + (-1881 = -110,431 \, \text{J}
\]
The negative sign indicates energy is released.

Rounding to the nearest whole number gives approximately \( 110432 \, \text{J} \).

Therefore, the correct answer is **C. 110,432 J**.