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:
- Heating the water from 25 °C to 100 °C.
- 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).