Question
How much heat energy is needed to change 95 g of water at 25 C into steam at 100 C? Use 4.2 J/g times C
for the specific heat of water. Use 2260 J/g for the heat of vaporization of water.
answers below:
2.4x 105 J
2.1x 105 J
1.8 × 104 J
3.0 × 104 J
for the specific heat of water. Use 2260 J/g for the heat of vaporization of water.
answers below:
2.4x 105 J
2.1x 105 J
1.8 × 104 J
3.0 × 104 J
Answers
GPT-4o mini
To calculate the total heat energy required to change 95 g of water at 25 °C into steam at 100 °C, we need to consider two steps:
1. Heating the water from 25 °C to 100 °C.
2. Vaporizing the water at 100 °C into steam.
### Step 1: Heating the water
The heat energy required to heat the water can be calculated using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
Where:
- \( Q \) = heat energy (in Joules)
- \( m \) = mass of the water (95 g)
- \( c \) = specific heat of water (4.2 J/g°C)
- \( \Delta T \) = change in temperature (final temperature - initial temperature)
Here, the final temperature is 100 °C and the initial temperature is 25 °C:
\[
\Delta T = 100 °C - 25 °C = 75 °C
\]
Now substituting the values into the formula:
\[
Q = 95 \, \text{g} \cdot 4.2 \, \text{J/g°C} \cdot 75 \, °C
\]
\[
Q = 95 \cdot 4.2 \cdot 75
\]
\[
Q = 95 \cdot 315
\]
\[
Q = 29925 \, \text{J}
\]
### Step 2: Vaporizing the water
Next, we calculate the heat energy required to vaporize the water using the heat of vaporization:
\[
Q_{\text{vaporization}} = m \cdot L
\]
Where:
- \( L \) = heat of vaporization of water = 2260 J/g
Now substituting the values:
\[
Q_{\text{vaporization}} = 95 \, \text{g} \cdot 2260 \, \text{J/g}
\]
\[
Q_{\text{vaporization}} = 214700 \, \text{J}
\]
### Total Heat Energy Required
Finally, to find the total heat energy needed, we sum the heat from both steps:
\[
Q_{\text{total}} = Q_{\text{heating}} + Q_{\text{vaporization}}
\]
\[
Q_{\text{total}} = 29925 \, \text{J} + 214700 \, \text{J}
\]
\[
Q_{\text{total}} = 244625 \, \text{J}
\]
Converting to scientific notation:
\[
Q_{\text{total}} = 2.44625 \times 10^5 \, \text{J} \approx 2.4 \times 10^5 \, \text{J}
\]
Thus, the amount of heat energy needed to change 95 g of water at 25 °C into steam at 100 °C is approximately:
**2.4 x 10^5 J** (the first answer option).
1. Heating the water from 25 °C to 100 °C.
2. Vaporizing the water at 100 °C into steam.
### Step 1: Heating the water
The heat energy required to heat the water can be calculated using the formula:
\[
Q = m \cdot c \cdot \Delta T
\]
Where:
- \( Q \) = heat energy (in Joules)
- \( m \) = mass of the water (95 g)
- \( c \) = specific heat of water (4.2 J/g°C)
- \( \Delta T \) = change in temperature (final temperature - initial temperature)
Here, the final temperature is 100 °C and the initial temperature is 25 °C:
\[
\Delta T = 100 °C - 25 °C = 75 °C
\]
Now substituting the values into the formula:
\[
Q = 95 \, \text{g} \cdot 4.2 \, \text{J/g°C} \cdot 75 \, °C
\]
\[
Q = 95 \cdot 4.2 \cdot 75
\]
\[
Q = 95 \cdot 315
\]
\[
Q = 29925 \, \text{J}
\]
### Step 2: Vaporizing the water
Next, we calculate the heat energy required to vaporize the water using the heat of vaporization:
\[
Q_{\text{vaporization}} = m \cdot L
\]
Where:
- \( L \) = heat of vaporization of water = 2260 J/g
Now substituting the values:
\[
Q_{\text{vaporization}} = 95 \, \text{g} \cdot 2260 \, \text{J/g}
\]
\[
Q_{\text{vaporization}} = 214700 \, \text{J}
\]
### Total Heat Energy Required
Finally, to find the total heat energy needed, we sum the heat from both steps:
\[
Q_{\text{total}} = Q_{\text{heating}} + Q_{\text{vaporization}}
\]
\[
Q_{\text{total}} = 29925 \, \text{J} + 214700 \, \text{J}
\]
\[
Q_{\text{total}} = 244625 \, \text{J}
\]
Converting to scientific notation:
\[
Q_{\text{total}} = 2.44625 \times 10^5 \, \text{J} \approx 2.4 \times 10^5 \, \text{J}
\]
Thus, the amount of heat energy needed to change 95 g of water at 25 °C into steam at 100 °C is approximately:
**2.4 x 10^5 J** (the first answer option).