Thermodynamics Change in Internal Energy?

Click For Summary
SUMMARY

The discussion focuses on calculating the heat transfer for a closed, rigid tank containing 2 kg of water transitioning from a two-phase liquid-vapor mixture at 70°C to saturated vapor at 120°C. The correct method involves using steam tables to determine the specific internal energies at both temperatures. The user initially calculated the heat transfer incorrectly as -466.5 kJ but later acknowledged the mistake, indicating the need for proper application of the internal energy change formula ΔU = Q - W, with W being zero in this scenario.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically internal energy changes.
  • Familiarity with steam tables for water at various temperatures.
  • Knowledge of the first law of thermodynamics, particularly the equation ΔU = Q - W.
  • Basic skills in performing calculations involving mass and specific internal energy.
NEXT STEPS
  • Study the use of steam tables for various substances, focusing on water.
  • Learn how to calculate heat transfer in closed systems using the first law of thermodynamics.
  • Explore the concept of phase changes in thermodynamics, particularly for water.
  • Practice problems involving internal energy calculations in rigid tanks.
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone involved in heat transfer calculations in closed systems.

Ricardeo Xavier
Messages
5
Reaction score
0

Homework Statement


A closed, rigid tank contains 2 kg of water, initially a two phase liquid–vapor mixture at T1 = 70°C. Heat transfer occurs until the tank contains only saturated vapor at T2 = 120C.
Determine the heat transfer for the process, in kJ.
answer choices:
3701kJ
119.4kJ
4835kJ
1558kJ

Homework Equations


Steam Tables
ΔU=Q-W(W=0)

The Attempt at a Solution


Since we know T1, we can use the tables to identify the specific internal energy of saturated liquid and gas and add them together to find u1. We can also do the same with T2 but neglecting the specific internal energy of fluid since it is only a saturated vapor at this point. Subtract u2 from u1 then multiply by the mass. Since it is a rigid tank we can also neglect work done. Here is my work:
Q=m(u2-u1)
Q=2kg(2529.3kJ/kg - (292.95kJ/kg+2469.6kJ/kg))
Q=-466.5kJ (this answer is not given as one of the choices so i am assuming that it is wrong)
 
Physics news on Phys.org
Nevermind i figured it out. I did this completely wrong
 

Similar threads

What is the change in internal energy after two processes?
  • · Replies 10 ·
Replies
10
Views
2K
Thermodynamics: Internal Energy, Heat and Work Problem
  • · Replies 1 ·
Replies
1
Views
2K
How Does Volume Affect Internal Energy in Thermodynamics?
  • · Replies 6 ·
Replies
6
Views
2K
Thermodynamics: find the change of internal energy, the work and Q
  • · Replies 10 ·
Replies
10
Views
2K
Internal Energy of an Ideal Gas
  • · Replies 4 ·
Replies
4
Views
3K
Change in internal energy of a cylinder
  • · Replies 8 ·
Replies
8
Views
2K
Adiabatic Expansion of Pressurized Air in a Piston-Cylinder Setup
  • · Replies 8 ·
Replies
8
Views
2K
Why does this solution assume it's a steady-flow system?
  • · Replies 6 ·
Replies
6
Views
2K
How Do You Calculate Heat Transfer in a Thermodynamics Piston Problem?
  • · Replies 5 ·
Replies
5
Views
3K
Is the change in internal energy really a state function?
  • · Replies 4 ·
Replies
4
Views
2K