When 1 mol of a gas burns at constant pressure, it produces 2424 J of heat and does 6 kJ of work.
How do i calculate delta E for this problem? I tried q+w but that didn't work for me. I got the answer wrong.
ΔE = q - w
ΔE is the change in internal energy of a system
q is the heat flowing into the system
w is the work being done by the system
The 2424 J is released to the surroundings. It should be assigned a negative sign. The 6 KJ is work done against the surroundings. That is energy released, too. Assign algebraic signs carefully and see what value for delta E you get.
Well, it seems like you're having a burning desire to find the right answer! Let me help you out.
To calculate the change in internal energy (delta E) for this problem, you can use the first law of thermodynamics, which states that the change in internal energy (delta E) is equal to the heat (q) plus the work (w) done on or by the system.
In this case, you've already been given the heat (q) as 2424 J and the work (w) as 6 kJ (which is 6000 J since 1 kJ = 1000 J).
So, delta E = q + w
= 2424 J + 6000 J
= 8424 J
So the change in internal energy (delta E) for this process is 8424 J.
Keep those chemical reactions burning, and don't hesitate to ask if you have any more questions!
In this problem, you are given q (heat) and w (work), and you need to calculate the change in internal energy (ΔE).
The equation you mentioned, ΔE = q + w, is the correct formula to calculate the change in internal energy. However, there might be some discrepancies in your calculations.
To calculate ΔE, you should use the following expression:
ΔE = q + w
Given that q = -2424 J (since heat is being released or lost) and w = -6000 J (since work is being done by the system against the surroundings), we can substitute these values into the equation:
ΔE = -2424 J + (-6000 J)
= -8424 J
Therefore, the change in internal energy (ΔE) for this problem is -8424 J. Please note that the negative sign indicates a decrease in internal energy.
To calculate the change in internal energy (ΔE) for this problem, you need to consider the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically, it can be expressed as:
ΔE = q - w
where q is the heat added to the system and w is the work done by the system.
In your case, you have been given:
- q = 2424 J (heat produced)
- w = 6 kJ = 6000 J (work done)
To ensure consistent units, it is important to convert kilojoules to joules by multiplying the value by 1000.
Substituting the given values into the equation, we can calculate ΔE as follows:
ΔE = q - w
ΔE = 2424 J - 6000 J
ΔE = -3576 J
The negative sign indicates that the system has lost energy. Thus, the change in internal energy (ΔE) for this process is -3576 J.