Energy into a rigid container>Find T2 and P2

  • Thread starter Thread starter boeing9379
  • Start date Start date
  • Tags Tags
    Energy
Click For Summary
SUMMARY

The discussion focuses on calculating the final temperature (T2) and pressure (P2) of air in a rigid container after a fan draws 1.5 kW of electricity for 30 minutes. The initial conditions are P1 = 1 atm and T1 = 27 degrees Celsius in a 60 m³ volume. The energy input into the system is calculated to be 2700.8 kJ, leading to a work done (W) of -2700.8 kJ due to the insulated nature of the container. The user seeks guidance on connecting energy changes with temperature and volume using the ideal gas law and internal energy equations.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of thermodynamic principles, specifically ΔE = Q - W
  • Familiarity with internal energy concepts (ΔU = mc(T2-T1))
  • Basic calculations involving energy conversions (kW to kJ)
NEXT STEPS
  • Study the relationship between internal energy and temperature changes in ideal gases.
  • Learn how to apply the First Law of Thermodynamics in closed systems.
  • Explore the concept of work done in thermodynamic processes, particularly in insulated systems.
  • Investigate the implications of rigid container behavior on pressure and temperature changes.
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineering, and physics who are working on problems related to energy transfer in closed systems, particularly those involving ideal gases and insulated environments.

boeing9379
Messages
3
Reaction score
0

Homework Statement


A Fan draws electricity at a rate of 1.5kW.
The fan is in a well insulated rigid container of air with volume 60 m^3.
Initial pressure and Initial Volume are: P1 = 1 atm, T1 = 27 degrees C.
After 30 minutes, assume Ideal Gas behavior, determine the final Temp and final Pressure.

Homework Equations



ΔE = Q - W
PV=nRT
ΔU=mc(T2-T1)?

The Attempt at a Solution



I calculated the energy into the system to be 2700.8 kJ after 30 minutes. Since there is no ΔKE or ΔPE, ΔE = ΔU = 2700.8 kJ. Since it is well insulated, Q = 0. So W = -2700.8 kJ.

I am having trouble linking the relationship between Energy and Temperature/Volume. I am not sure what other equations to use, so I am making no progress on this problem.
 
Last edited:
Physics news on Phys.org
Does anyone have some advice. I will take anything that points me in the right direction, lol.
 

Similar threads

Energy Balance for closed rigid container
  • · Replies 11 ·
Replies
11
Views
9K
Thermodynamics energy balance for control volume
  • · Replies 1 ·
Replies
1
Views
2K
Calculating entropy for an adiabatic system
  • · Replies 11 ·
Replies
11
Views
3K
Thermodynamics Energy transfer question
  • · Replies 10 ·
Replies
10
Views
2K
Applying first law of thermo to a closed spring piston
  • · Replies 2 ·
Replies
2
Views
4K
Thermodynamics -- Internal Energy
  • · Replies 15 ·
Replies
15
Views
6K
Thermodynamics Piston-Cylinder Question
  • · Replies 4 ·
Replies
4
Views
12K
Determining change in internal energy and enthelpy
  • · Replies 12 ·
Replies
12
Views
4K
Thermodynamics - Two gases in a container
  • · Replies 14 ·
Replies
14
Views
9K
Thermodynamics Energy Balance: Solving for Work Input in an Isothermal Process
  • · Replies 23 ·
Replies
23
Views
5K