The amount of energy released by burning a fuel source, measured in energy per mass, is called fuel value. If all the energy obtained from burning 1.3 pounds of butane (fuel value 10.85 kcal/g) is used to heat 122 kg of water at an initial temperature of 20.9 degC, what is the final temperature?
10.85 kcal/g x 1.3 pounds x (454 g/pound) = ?? kcal.
??kcal = mass H2O x specific heat water x (Tfinal - Tinitial)
Solve for Tfinal.
A gas at 65oC occupies 4.22 L At what Celsius temperature will the volume be 3.87 L assuming the same pressure?
(V1/T1) = (V2/T2)
T must be in Kelvin.
To find the final temperature of the water, we can use the principle of conservation of energy. The energy released by burning 1.3 pounds of butane will be used to heat the water.
Step 1: Convert the mass of butane from pounds to grams.
1.3 pounds ≈ 589.67 grams
Step 2: Calculate the total energy released by burning the butane.
Energy released = mass of butane x fuel value
Energy released = 589.67 g x 10.85 kcal/g
Step 3: Convert the energy released from kcal to joules.
1 kcal ≈ 4.184 kJ
Energy released = (589.67 g x 10.85 kcal/g) x 4.184 kJ/kcal
Step 4: Calculate the energy required to heat the water.
Energy required = mass of water x specific heat x change in temperature
Given:
Mass of water = 122 kg
Specific heat of water = 4.186 J/g°C (approximately)
First, convert the mass of water from kg to grams:
122 kg = 122,000 grams
Energy required = 122,000 g x 4.186 J/g°C x (final temperature - 20.9°C)
Step 5: Equate the energy released to the energy required and solve for the final temperature.
(589.67 g x 10.85 kcal/g x 4.184 kJ/kcal) = (122,000 g x 4.186 J/g°C x (final temperature - 20.9°C))
Solve the equation for the final temperature:
(final temperature - 20.9°C) = (589.67 g x 10.85 kcal/g x 4.184 kJ/kcal) / (122,000 g x 4.186 J/g°C)
Finally, add 20.9°C to the result obtained to find the final temperature.