Certainly! Let's go through the detailed steps to determine how many kilowatt-hours (kWh) of energy are required to heat 75 gallons of water by 25°C.
### Step-by-Step Solution:
1. Convert Volume of Water from Gallons to Liters:
- Given volume in gallons: 75 gal
- Conversion factor: \( 1 \text{ gallon} = 3.78541 \text{ liters} \)
[tex]\[
\text{Volume in liters} = 75 \text{ gal} \times 3.78541 \text{ L/gal} = 283.91 \text{ L}
\][/tex]
2. Calculate the Mass of Water:
- Density of water: \( 1 \text{ liter of water} = 1000 \text{ grams} \)
[tex]\[
\text{Mass in grams} = 283.91 \text{ L} \times 1000 \text{ g/L} = 283,906 \text{ g}
\][/tex]
3. Calculate the Energy Required in Joules:
- Specific heat capacity of water: \( 4.186 \text{ J/(g°C)} \)
- Temperature change: \( \Delta T = 25 \text{ °C} \)
[tex]\[
\text{Energy (Joules)} = \text{mass (g)} \times \text{specific heat capacity} \times \Delta T
\][/tex]
[tex]\[
\text{Energy (J)} = 283,906 \text{ g} \times 4.186 \text{ J/(g°C)} \times 25 \text{ °C} = 29,710,737 \text{ J}
\][/tex]
4. Convert the Energy from Joules to Kilowatt-Hours:
- Conversion factor: \( 1 \text{ Joule} = 2.77778 \times 10^{-7} \text{ kWh} \)
[tex]\[
\text{Energy (kWh)} = 29,710,737 \text{ J} \times 2.77778 \times 10^{-7} \text{ kWh/J} = 8.25 \text{ kWh}
\][/tex]
### Final Answer:
To heat 75 gallons of water by 25°C, you would need 8.3 kWh of energy (rounded to two significant figures).
