Question
How much heat energy is needed to change 95 g of water at 25°C into steam at 100°C ? Use 4.2 Jg⋅°C for the specific heat of water. Use 2260 J/g for the heat of vaporization of water.(1 point) Responses 2.1×105 J 2 point 1 times 10 to the 5th power joule 2.4×105 J 2 point 4 times 10 to the 5th power joule 1.8×104 J 1 point 8 times 10 to the 4th power joule 3.0×104 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 to steam.**
### Step 1: Heating the Water
The formula for calculating the heat required to warm a substance is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( Q_1 \) = heat required (in joules)
- \( m \) = mass (in grams)
- \( c \) = specific heat capacity (J/g°C)
- \( \Delta T \) = change in temperature (°C)
**Values:**
- \( m = 95 \, g \)
- \( c = 4.2 \, J/g°C \)
- The change in temperature \( \Delta T = 100°C - 25°C = 75°C \)
**Calculation:**
\[ Q_1 = 95 \, g \cdot 4.2 \, J/g°C \cdot 75 \, °C \]
\[ Q_1 = 95 \cdot 4.2 \cdot 75 \]
\[ Q_1 = 95 \cdot 315 \]
\[ Q_1 = 29925 \, J \]
### Step 2: Vaporizing the Water
The heat required to vaporize water is given by:
\[ Q_2 = m \cdot L_v \]
where:
- \( Q_2 \) = heat required for vaporization (in joules)
- \( m \) = mass (in grams)
- \( L_v \) = heat of vaporization (J/g)
**Values:**
- \( L_v = 2260 \, J/g \)
**Calculation:**
\[ Q_2 = 95 \, g \cdot 2260 \, J/g \]
\[ Q_2 = 214700 \, J \]
### Total Heat Energy
Now, the total heat energy \( Q_{total} \) required is:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 29925 \, J + 214700 \, J \]
\[ Q_{total} = 244625 \, J \]
### Final Answer
In scientific notation, this is:
\[ Q_{total} \approx 2.45 \times 10^5 \, J \]
Thus, the closest response from the options provided is **2.4×10^5 J.**
1. **Heating the water from 25°C to 100°C.**
2. **Vaporizing the water at 100°C to steam.**
### Step 1: Heating the Water
The formula for calculating the heat required to warm a substance is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \( Q_1 \) = heat required (in joules)
- \( m \) = mass (in grams)
- \( c \) = specific heat capacity (J/g°C)
- \( \Delta T \) = change in temperature (°C)
**Values:**
- \( m = 95 \, g \)
- \( c = 4.2 \, J/g°C \)
- The change in temperature \( \Delta T = 100°C - 25°C = 75°C \)
**Calculation:**
\[ Q_1 = 95 \, g \cdot 4.2 \, J/g°C \cdot 75 \, °C \]
\[ Q_1 = 95 \cdot 4.2 \cdot 75 \]
\[ Q_1 = 95 \cdot 315 \]
\[ Q_1 = 29925 \, J \]
### Step 2: Vaporizing the Water
The heat required to vaporize water is given by:
\[ Q_2 = m \cdot L_v \]
where:
- \( Q_2 \) = heat required for vaporization (in joules)
- \( m \) = mass (in grams)
- \( L_v \) = heat of vaporization (J/g)
**Values:**
- \( L_v = 2260 \, J/g \)
**Calculation:**
\[ Q_2 = 95 \, g \cdot 2260 \, J/g \]
\[ Q_2 = 214700 \, J \]
### Total Heat Energy
Now, the total heat energy \( Q_{total} \) required is:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 29925 \, J + 214700 \, J \]
\[ Q_{total} = 244625 \, J \]
### Final Answer
In scientific notation, this is:
\[ Q_{total} \approx 2.45 \times 10^5 \, J \]
Thus, the closest response from the options provided is **2.4×10^5 J.**