Question

The amount of energy needed to boil 75 g of water, if initially at 60C, is blank.
(1 point)
Answers below:
169,500 J

37,590 J

182,040 J

12,540 J

Answer 1

To calculate the amount of energy needed to boil 75 g of water that is initially at 60°C, we can use the formula:

\[
Q = mc\Delta T + mL
\]

Where:
- \( Q \) is the total energy in joules (J)
- \( m \) is the mass of the water in grams (g)
- \( c \) is the specific heat capacity of water (approximately \( 4.18 \, \text{J/g°C} \))
- \( \Delta T \) is the change in temperature in degrees Celsius (°C)
- \( L \) is the latent heat of vaporization of water (approximately \( 2260 \, \text{J/g} \))

1. **Calculate the change in temperature (\( \Delta T \))**:
- The boiling point of water is 100°C.
- The initial temperature is 60°C.
- Therefore, \( \Delta T = 100°C - 60°C = 40°C \).

2. **Calculate the energy needed to heat the water to boiling**:
\[
Q_1 = mc\Delta T = (75 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (40°C)
\]
\[
Q_1 = 75 \times 4.18 \times 40 = 12,540 \, \text{J}
\]

3. **Calculate the energy needed to convert the water to steam**:
\[
Q_2 = mL = (75 \, \text{g}) \times (2260 \, \text{J/g})
\]
\[
Q_2 = 75 \times 2260 = 169,500 \, \text{J}
\]

4. **Total energy required**:
\[
Q = Q_1 + Q_2 = 12,540 \, \text{J} + 169,500 \, \text{J} = 182,040 \, \text{J}
\]

So the amount of energy needed to boil 75 g of water initially at 60°C is **182,040 J**.