Calculate the heat absorbed

In summary, an ideal gas with a γ value of 1.4 was expanded under constant pressure, resulting in 80 Joules of work done. The question asks to calculate the absorbed heat and increase in internal energy of the system. The solution involves using the equation w = PΔV and the first law of thermodynamics, where work is equal to the change in energy plus heat. By dividing the equations for heat and change in energy, the value for γ can be obtained, allowing for the calculation of heat and change in energy with two equations and two unknowns.
  • #1
dk_ch
44
0

Homework Statement



An ideal gas(γ=1.4) was expanded under constant pressure. The work done was 80 Joule.Calculate the heat absorbed and increase in internal energy of the system.
 
Physics news on Phys.org
  • #2
Well, what's the equation for a gas undergoing expansion under constant pressure?
[itex]w= P \int dV[/itex]
 
  • #3
Feodalherren said:
Well, what's the equation for a gas undergoing expansion under constant pressure?
[itex]w= P \int dV[/itex]

what is the solution?
 
  • #4
I'm not just going to give you the solution. You won't learn anything. You need to show some work and initiative and I can guide you through it.

Start by thinking about this: is the change in energy path dependent or path independent?
 
  • #5
Feodalherren said:
I'm not just going to give you the solution. You won't learn anything. You need to show some work and initiative and I can guide you through it.

Start by thinking about this: is the change in energy path dependent or path independent?

W = Q-ΔU is known from from first law of thermodynamics and w is given here . Now how shall i proceed using γ=1.4 ? how to apply your equation ?
please help.
 
Last edited:
  • #6
Is this calculus based physics or algebra based?

See this is why you need to show some work, I have no idea at what level you're doing this. If the integral is what's confusing you then simply change it to: [itex]w = P \Delta V[/itex].

You want to find [itex]Q_{in}[/itex] for a constant pressure problem.

[itex]\Delta E = W + Q[/itex]

We know that w= -80J. Because the system did work on the piston it has to be negative - energy left the system.

[itex]\Delta E + 80 = Q[/itex]

[itex]\gamma = 1.4[/itex] Therefore you know it's a diatomic gas.

And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

[itex]Q = nC_{p} \Delta T[/itex]

and [itex]\Delta E = nC_{v} \Delta T[/itex]

Edit: It's ambiguous if they mean that the work done was by the system or on the system. Is this the whole original question?
 
Last edited:
  • #7
Feodalherren said:
And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

[itex]Q = nC_{p} \Delta T[/itex]

and [itex]\Delta E = nC_{v} \Delta T[/itex]

Divide these two equations and you will get Q/ΔE = cp/cv = γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.
 
  • #8
CAF123 said:
Divide these two equations and you will get Q/ΔE = cp/cv = γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.
Ah! Genius. I would never have thought of that :).
 
  • #9
CAF123 said:
Divide these two equations and you will get Q/ΔE = cp/cv = γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.

well done. Thanks to both of you.
 

1. How do you calculate the heat absorbed?

The formula for calculating heat absorbed is Q = mcΔT, where Q is the heat absorbed, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

2. What is specific heat capacity?

Specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius.

3. Can you give an example of calculating heat absorbed?

Sure, for example, if we have a 500-gram block of iron with a specific heat capacity of 0.45 J/g°C and it's heated from 20°C to 50°C, the heat absorbed would be Q = (500 g)(0.45 J/g°C)(50°C - 20°C) = 9,000 J.

4. What units are used for heat absorbed?

The units for heat absorbed are Joules (J) in the metric system and calories (cal) in the imperial system.

5. How is heat absorbed different from heat released?

Heat absorbed is the amount of heat energy that is gained by an object, while heat released is the amount of heat energy that is lost by an object. They are both measured in the same units (Joules or calories) but have opposite effects on the temperature of the object.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
899
  • Introductory Physics Homework Help
Replies
1
Views
867
  • Introductory Physics Homework Help
Replies
12
Views
726
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
626
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
895
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top