3. One calorie is the energy required to raise one gram of water one degree Celsius. Since we use the joule as the standard measurement of energy, 1 calorie = 4.184 J. If a pot is placed on the stove with 2.4 liters of water at a temperature of 27°C, how many joules are required to bring the water to its boiling point? You may neglect heat loss to the environment. If this occurs in a period of 11 minutes, what is the average rate of heat transfer in joules per second? This is for the entire 2.4 liters of water. What if we want to know the heat transfer rate for just 1 liter of water? Assuming this occurs on an electric stove and the pan completely covers the burner, what type of energy transfer is occurring?
2.4LH2O*1000gH2O/1LH2O=2400gH2O
27cal*4.184J/99.97cal=1.13J
2400*1.13J=2.712*10^4J/g
I don't know if i'm doing this right?
To calculate the amount of energy required to bring the water to its boiling point, you can use the specific heat capacity formula:
Q = mcΔT
where:
Q is the amount of energy required (in joules)
m is the mass of water (in grams)
c is the specific heat capacity of water (in joules/gram °C)
ΔT is the change in temperature (in °C)
In this case, you have 2400 grams of water (2.4 liters * 1000 grams/liter) at a temperature of 27°C. Since you want to bring it to its boiling point, the change in temperature is ΔT = 100°C (assuming water boils at 100°C).
Now, you can substitute the values into the formula:
Q = (2400 g) * (4.184 J/g°C) * (100°C) =
= 1,002,816 J
Therefore, you would require approximately 1,002,816 joules of energy to bring the 2.4 liters of water to its boiling point.
To find the average rate of heat transfer in joules per second, you need to divide the total energy by the time taken:
Average rate of heat transfer = (1,002,816 J) / (11 minutes * 60 seconds/minute) =
≈ 1525.6 J/s
So, the average rate of heat transfer for the entire 2.4 liters of water is approximately 1525.6 joules per second.
If you want to know the heat transfer rate for just 1 liter of water, you can divide the total energy by the volume of water:
Heat transfer rate for 1 liter = (1,002,816 J) / (2.4 liters) =
≈ 417,840 J/liter
Therefore, the heat transfer rate for just 1 liter of water is approximately 417,840 joules per liter.
Since the pan completely covers the burner on an electric stove, the type of energy transfer occurring is primarily conduction. Heat is being transferred from the electric burner to the pan, and then from the pan to the water through direct contact.