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Sagot :
To solve this problem, we need to find how much energy is required to heat 75 gallons of water by 25 degrees Celsius, and express that energy in kilowatt-hours.
Here’s a step-by-step solution:
1. Convert the volume of water from gallons to liters.
- The given volume is 75 gallons.
- We know that 1 gallon is approximately 3.785 liters.
- Therefore, the volume in liters is:
[tex]\[ 75 \, \text{gal} \times 3.785 \, \text{L/gal} = 283.875 \, \text{L} \][/tex]
2. Determine the mass of the water.
- The density of water is approximately 1 kg/L.
- Therefore, the mass of the water in kilograms is equal to its volume in liters:
[tex]\[ 283.875 \, \text{L} \times 1 \, \text{kg/L} = 283.875 \, \text{kg} \][/tex]
3. Calculate the energy required in Joules.
- We use the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the water (283.875 kg),
- \( c \) is the specific heat capacity of water (4.186 J/g°C),
- \( \Delta T \) is the temperature change (25°C).
Convert the mass to grams (1 kg = 1000 g):
[tex]\[ 283.875 \, \text{kg} \times 1000 \, \text{g/kg} = 283875 \, \text{g} \][/tex]
Now, calculate the energy:
[tex]\[ Q = 283875 \, \text{g} \times 4.186 \, \text{J/g°C} \times 25 \, \text{°C} = 29,707,521.25 \, \text{J} \][/tex]
4. Convert the energy from Joules to kilowatt-hours.
- We know that 1 kilowatt-hour (kWh) is equal to \( 3.6 \times 10^6 \) Joules.
- Therefore, the energy in kilowatt-hours is:
[tex]\[ \frac{29,707,521.25 \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} = 8.252 \times 10^{-3} \, \text{kWh} \][/tex]
5. Express the answer to two significant figures.
- Rounding the value to two significant figures, we get:
[tex]\[ 0.01 \, \text{kWh} \][/tex]
Therefore, the energy necessary to heat 75 gallons of water by 25 degrees Celsius is 0.01 kilowatt-hours.
Here’s a step-by-step solution:
1. Convert the volume of water from gallons to liters.
- The given volume is 75 gallons.
- We know that 1 gallon is approximately 3.785 liters.
- Therefore, the volume in liters is:
[tex]\[ 75 \, \text{gal} \times 3.785 \, \text{L/gal} = 283.875 \, \text{L} \][/tex]
2. Determine the mass of the water.
- The density of water is approximately 1 kg/L.
- Therefore, the mass of the water in kilograms is equal to its volume in liters:
[tex]\[ 283.875 \, \text{L} \times 1 \, \text{kg/L} = 283.875 \, \text{kg} \][/tex]
3. Calculate the energy required in Joules.
- We use the formula:
[tex]\[ Q = m \cdot c \cdot \Delta T \][/tex]
where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the water (283.875 kg),
- \( c \) is the specific heat capacity of water (4.186 J/g°C),
- \( \Delta T \) is the temperature change (25°C).
Convert the mass to grams (1 kg = 1000 g):
[tex]\[ 283.875 \, \text{kg} \times 1000 \, \text{g/kg} = 283875 \, \text{g} \][/tex]
Now, calculate the energy:
[tex]\[ Q = 283875 \, \text{g} \times 4.186 \, \text{J/g°C} \times 25 \, \text{°C} = 29,707,521.25 \, \text{J} \][/tex]
4. Convert the energy from Joules to kilowatt-hours.
- We know that 1 kilowatt-hour (kWh) is equal to \( 3.6 \times 10^6 \) Joules.
- Therefore, the energy in kilowatt-hours is:
[tex]\[ \frac{29,707,521.25 \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} = 8.252 \times 10^{-3} \, \text{kWh} \][/tex]
5. Express the answer to two significant figures.
- Rounding the value to two significant figures, we get:
[tex]\[ 0.01 \, \text{kWh} \][/tex]
Therefore, the energy necessary to heat 75 gallons of water by 25 degrees Celsius is 0.01 kilowatt-hours.
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