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.
•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.
1 answer