How much energy does it take to raise the temperature of 20 gallons of water from 5C to 300 F? Round answer to the nearest kJ.

Answer 1

First, let's convert gallons to liters:
1 gallon = 3.78541 liters
20 gallons = 20 * 3.78541 liters = 75.7082 liters

Next, let's convert the temperatures to the same unit to find the temperature difference:
1C = 1K, so 5C = 5 + 273.15 K = 278.15 K
Convert 300F to Celsius:
300F = (300 - 32) * 5/9 = 148.8889 C
148.8889 C = 148.8889 K + 273.15 K = 422.0389 K

Temperature difference: (422.0389 - 278.15) K = 143.8889 K

Now, let's find the energy needed to warm up the water:
Specific heat capacity of water = 4.186 J/gK
1 liter of water = 1 kg = 1000 g
Energy = mass * specific_heat * temperature_difference
Energy = 75.7082 L * 1000 g/L * 4.186 J/gK * 143.8889 K

Energy = 45406047.6287 J

Since 1 kJ = 1000 J, we can convert the energy:
Energy = 45406047.6287 J / 1000 J/kJ = 45406.047 kJ

Rounding to the nearest kJ, the energy required to raise the temperature of 20 gallons of water from 5C to 300 F is 45,406 kJ.