Question

How much paraffin oil would be needed to heat the same volume of water?

Answer 1

To determine how much paraffin oil would be needed to heat the same volume of water, we need to know the specific heat capacities of both water and paraffin oil, as well as the temperature change desired.

1. **Specific Heat Capacity**:
- Water: Approximately \( 4.18 \, \text{J/g°C} \)
- Paraffin Oil: Approximately \( 2.1 \, \text{J/g°C} \)

2. **Equation for Heat Transfer**:
The heat gained or lost can be calculated using the equation:
\[
Q = mc\Delta T
\]
where:
\( Q \) is the heat energy (in joules),
\( m \) is the mass (in grams),
\( c \) is the specific heat capacity (in J/g°C),
\( \Delta T \) is the change in temperature (in °C).

3. **Example Calculation**:
Let's assume we want to heat 1 liter (1000 grams) of water from an initial temperature of 20°C to a final temperature of 100°C. Here, the temperature change (\( \Delta T \)) is \( 100°C - 20°C = 80°C \).

For water:
\[
Q_{\text{water}} = (1000 \, g)(4.18 \, \text{J/g°C})(80 \, °C) = 334,400 \, J
\]

To achieve the same amount of energy using paraffin oil, we need to rearrange the equation:
\[
Q_{\text{oil}} = m_{\text{oil}}c_{\text{oil}}\Delta T_{\text{oil}}
\]
Assuming we want to heat the same amount of water by the same temperature \( \Delta T \), we can set \( Q_{\text{oil}} = Q_{\text{water}} = 334,400 \, J \).

Now, let’s assume we want to heat the paraffin oil by the same \( \Delta T = 80°C \):
\[
334,400 \, J = m_{\text{oil}} (2.1 \, \text{J/g°C})(80 \, °C)
\]

Solving for \( m_{\text{oil}} \):
\[
334,400 = m_{\text{oil}} \cdot 168 \quad (\text{since } 2.1 \times 80 = 168)
\]
\[
m_{\text{oil}} = \frac{334,400}{168} \approx 1,993 \, g
\]

4. **Conclusion**:
To heat the same volume of 1 liter of water from 20°C to 100°C, you would need approximately 1,993 grams of paraffin oil, assuming it is heated by the same temperature difference.