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Bot To find out how much energy it takes to raise the temperature of 20 gallons of water from 5°C to 300°F, we will use the formula:
Q = mcΔT,
where Q is the energy required (in joules),
m is the mass of the water (in kilograms),
c is the specific heat capacity of water (4.18 J/g°C = 4.18 kJ/kg°C),
and ΔT is the change in temperature (in degrees Celsius).
First, we'll need to convert gallons of water to kilograms and Fahrenheit to Celsius.
1 gallon of water = 3.78541 liters = 3.78541 kg (1 liter of water weighs 1 kg)
20 gallons = 20 * 3.78541 kg = 75.7082 kg
Next, we'll convert 300°F to Celsius:
(300°F - 32) * 5/9 = 149°C
Change in temperature (ΔT):
149°C - 5°C = 144°C
Now, we'll calculate the energy required:
Q = m * c * ΔT
Q = 75.7082 kg * 4.18 kJ/kg°C * 144°C
Q ≈ 45681.386 kJ
Rounding to the nearest kJ, the answer is 45,681 kJ.