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.
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.
To calculate the energy required to raise the temperature of water, you need to use the heat capacity of water. The heat capacity of water is approximately 4.184 J/g·°C.
First, convert the volume of water from gallons to grams. We know that 1 gallon is approximately equal to 3.78541 liters, and since the density of water is 1 g/mL, we can multiply the volume in gallons by 3.78541 to get the volume in liters. Finally, multiply the volume in liters by 1000 to convert it to grams.
Volume in liters = 20 gallons * 3.78541 (L/gallon) ≈ 75.7082 L
Volume in grams = 75.7082 L * 1000 (g/L) ≈ 75708.2 g
Next, calculate the difference in temperature between the initial and final temperatures in Celsius:
ΔT = 300°F - 5°C ≈ 295°C
Finally, use the heat capacity of water to calculate the energy required:
Energy (in J) = heat capacity (in J/g·°C) * mass (in g) * temperature change (in °C)
Energy (in J) = 4.184 J/g·°C * 75708.2 g * 295°C ≈ 90,230,099 J
To convert this energy to kJ, divide by 1000:
Energy (in kJ) ≈ 90,230,099 J / 1000 ≈ 90,230 kJ
Therefore, it takes approximately 90,230 kJ of energy to raise the temperature of 20 gallons of water from 5°C to 300°F.
To calculate the energy required to raise the temperature of water, we can use the equation:
Q = m * c * ΔT
Where:
Q is the energy (in joules)
m is the mass of water (in kilograms)
c is the specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
First, let's convert the given values to the appropriate units:
20 gallons of water = 20 * 3.78541 liters = 75.7082 liters
1 liter of water = 1 kilogram (approximately)
ΔT = 300°F - 5°C = 300°F - 5 * (9/5)°C = 300°F - 9°F = 291°F
Now, we need to convert the change in temperature to degrees Celsius:
ΔT = 291°F * (5/9)°C/°F = 161.67°C
Next, let's find the specific heat capacity of water. The specific heat capacity of water is approximately 4.186 joules per gram per degree Celsius, which we can convert to kilograms:
c = 4.186 J/g°C * 1 kg/1000 g = 0.004186 J/g°C
Now, we can calculate the energy required:
Q = m * c * ΔT = 75.7082 kg * 0.004186 J/g°C * 161.67°C ≈ 50.01 kJ
Therefore, it would take approximately 50.01 kilojoules of energy to raise the temperature of 20 gallons of water from 5°C to 300°F.