If all the energy obtained from burning 1.23 pounds of propane (fuel value is 11.09 kcal/g) is used to heat 125.0 kg of water at an initial temperature of 21.8 °C, what is the final temperature?
Hint:
The heat, q, released by burning propane is absorbed by the water. Use the following formula to find the temperature change, ÄT,
q = m*c*(Change in temperature)
Where m is mass and c is specific heat
I've done it several different ways and I keep getting 21.8 for an answer and it says that is wrong.
To solve this problem, we can use the formula for heat transfer:
q = m * c * ΔT
Where:
q is the heat transferred (in calories)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in calories per gram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
Given information:
Mass of water, m = 125.0 kg = 125000 grams
Initial temperature, Ti = 21.8 °C
Fuel value of propane, 11.09 kcal/g
First, we need to calculate the heat released by burning 1.23 pounds of propane:
Conversion factor: 1 pound = 453.592 grams
Mass of propane, m_propane = 1.23 pounds * 453.592 grams/pound = 558.096 grams
Heat released, q_propane = m_propane * fuel value = 558.096 g * 11.09 kcal/g
= 6185.44144 kcal
Now, we can equate the heat released by burning propane to the heat absorbed by the water:
q_propane = m_water * c * ΔT
Rearranging the equation to solve for ΔT:
ΔT = q_propane / (m_water * c)
Plugging in the known values:
ΔT = 6185.44144 kcal / (125000 g * c)
The specific heat capacity of water is approximately 1 calorie/gram/°C. Therefore:
ΔT = 6185.44144 kcal / (125000 g * 1 cal/g/°C)
= 0.04948353 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Tf = Ti + ΔT
Tf = 21.8 °C + 0.04948353 °C
So, the final temperature of the water is approximately 21.8 °C.
If you have been getting the same answer and it is marked as incorrect, there may be a mistake in the input values or calculations given. Double-check the numbers used in the formula above and make sure to perform accurate calculations to find the correct answer.
To find the final temperature, we can use the formula q = m * c * ΔT, where q is the heat released by burning propane, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.
First, let's calculate the heat released by burning propane. The energy obtained from burning 1.23 pounds (0.558 kg) of propane is:
Energy = fuel value * mass
Energy = 11.09 kcal/g * 0.558 kg
Energy = 6.17222 kcal
Since 1 kcal is equal to 4.18 kJ, the energy obtained from burning propane is:
Energy = 6.17222 kcal * 4.18 kJ/kcal
Energy = 25.7524376 kJ
Next, we need to calculate the mass of water:
mass of water = 125.0 kg
Now, let's calculate the change in temperature (ΔT) using the formula:
ΔT = energy / (m * c)
where m is the mass of water, and c is the specific heat of water.
ΔT = (25.7524376 kJ) / (125.0 kg * 4.18 kJ/kg·°C)
ΔT = 0.1559 °C
Finally, to find the final temperature, we can add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 21.8 °C + 0.1559 °C
Final temperature = 21.9559 °C
Therefore, the final temperature is approximately 21.96 °C.
Change 1.23 pounds to grams.
1.23 pounds x (453.6 g/lb) = ?
?g x 11,090 cal/g = calories of heat produced = y
y = mass H2O x specific heat H2O x (Tfinal-Tinitial).
I would use 125,000 for mass H2O. I get something like 70 C. You know it can't be 21.8. That's the starting temperature and you're adding bunches of heat. T must go up.