Question
How much heat energy is needed to change 95 g of water at 25°C into steam at 100°C? Use 4.2 for the
specific heat of water. Use 2260 J/g for the heat of vaporization of water. (1 point)
2.1 x 10 J
1.8 x 10 J
2.4 x 10 J
3.0 x 10 J
specific heat of water. Use 2260 J/g for the heat of vaporization of water. (1 point)
2.1 x 10 J
1.8 x 10 J
2.4 x 10 J
3.0 x 10 J
Answers
GPT-4o mini
To find the total heat energy needed 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:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( Q_1 \) = heat energy (in joules)
- \( m \) = mass of water (95 g)
- \( c \) = specific heat of water (4.2 J/g°C)
- \( \Delta T \) = change in temperature (final temperature - initial temperature = 100°C - 25°C = 75°C)
Substituting the values:
\[ Q_1 = 95 \, \text{g} \cdot 4.2 \, \text{J/g°C} \cdot 75 \, \text{°C} \]
\[ Q_1 = 95 \cdot 4.2 \cdot 75 \]
\[ Q_1 = 23925 \, \text{J} \]
\[ Q_1 = 2.39 \times 10^4 \, \text{J} \]
### Step 2: Vaporizing the water
The heat energy required for the phase change (vaporization) can be calculated using the formula:
\[ Q_2 = m \cdot L_v \]
where:
- \( Q_2 \) = heat energy for vaporization (in joules)
- \( m \) = mass of water (95 g)
- \( L_v \) = heat of vaporization (2260 J/g)
Substituting the values:
\[ Q_2 = 95 \, \text{g} \cdot 2260 \, \text{J/g} \]
\[ Q_2 = 214700 \, \text{J} \]
\[ Q_2 = 2.15 \times 10^5 \, \text{J} \]
### Total Heat Energy
Now, we sum the heat energies from both steps:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 23925 \, \text{J} + 214700 \, \text{J} \]
\[ Q_{total} = 238625 \, \text{J} \]
\[ Q_{total} = 2.39 \times 10^5 \, \text{J} \]
Considering the provided options and converting \( 2.39 \times 10^5 \) J into a suitable scientific format, we find:
The answer is closest to **2.4 x 10^5 J**.
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:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( Q_1 \) = heat energy (in joules)
- \( m \) = mass of water (95 g)
- \( c \) = specific heat of water (4.2 J/g°C)
- \( \Delta T \) = change in temperature (final temperature - initial temperature = 100°C - 25°C = 75°C)
Substituting the values:
\[ Q_1 = 95 \, \text{g} \cdot 4.2 \, \text{J/g°C} \cdot 75 \, \text{°C} \]
\[ Q_1 = 95 \cdot 4.2 \cdot 75 \]
\[ Q_1 = 23925 \, \text{J} \]
\[ Q_1 = 2.39 \times 10^4 \, \text{J} \]
### Step 2: Vaporizing the water
The heat energy required for the phase change (vaporization) can be calculated using the formula:
\[ Q_2 = m \cdot L_v \]
where:
- \( Q_2 \) = heat energy for vaporization (in joules)
- \( m \) = mass of water (95 g)
- \( L_v \) = heat of vaporization (2260 J/g)
Substituting the values:
\[ Q_2 = 95 \, \text{g} \cdot 2260 \, \text{J/g} \]
\[ Q_2 = 214700 \, \text{J} \]
\[ Q_2 = 2.15 \times 10^5 \, \text{J} \]
### Total Heat Energy
Now, we sum the heat energies from both steps:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 23925 \, \text{J} + 214700 \, \text{J} \]
\[ Q_{total} = 238625 \, \text{J} \]
\[ Q_{total} = 2.39 \times 10^5 \, \text{J} \]
Considering the provided options and converting \( 2.39 \times 10^5 \) J into a suitable scientific format, we find:
The answer is closest to **2.4 x 10^5 J**.