1. First Law of Thermodynamics
Chapter Five:
First Law of Thermodynamics

2. First Law of thermodynamics
States that:
“Energy cannot be created or destroyed BUT it can be changed from one form to another”

3. Therefore:
The first law of thermodynamics is an expression of the conservation of energy principle.
Energy can cross the boundaries of a closed system in the form of heat or work.

4. Energy Balance

5. Conservation of Mass
“Mass cannot be created or destroyed BUT it can be changed from one form to another”

6. Or: Min-Mout=ΔMsystem
Mass Balance
Mass
Mass
Mass M M
Or: Min-Mout=ΔMsystem

8. First Law for Closed System
For the closed system shown below, the conservation of energy principle or the first law of thermodynamics is expressed as
Heat
Work z
Closed
System
Reference Plane, z = 0 or
For stationary closed system:

9. = 0

10. 1st Law of thermodynamics
Tutorials
1st Law of thermodynamics

11. Problem 1
On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will the room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed.

12. Answer
Warmer. Because energy is added to the room air in the form of electrical work.

13. Problem 2
A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 65 Btu of heat is transferred to the water, and heat losses from the side walls amount to 8 Btu. The piston rises as a result of evaporation, and 5 Btu of boundary work is done. Determine the change in the energy of the water for this process.

15. Problem 3
A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at l00 kPa. Initially, three quarters of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 110-V source, and a current of 8 A flows through the resistor when the switch is turned on. Determine how long it will take to vaporize all the liquid in the tank. Also, show the process on a T-υ diagram with respect to saturation lines.
H2O
V = const. We

16. Answer

18. v T 1 2

19. Problem 4
Is it possible to compress an ideal gas isothermally in an adiabatic piston-cylinder device? Explain.

20. Answer
No, it isn't. This is because the first law relation Q - W = U reduces to W = 0 in this case since the system is adiabatic (Q = 0) and U = 0 for the isothermal processes of ideal gases. Therefore, this adiabatic system cannot receive any net work at constant temperature.

21. Problem 5
A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa and 25°C. The mass of the piston is such that 500 kPa of pressure is required to raise it. How much heat must be transferred to the helium before the piston starts rising? He
100 kPa
25C Q
500 kPa

24. Problem 6
A 4-m X 5-m X 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10,000 kJ/h, and a 100-W fan is used to distribute the warm air in the room. The rate of heat loss from the room is estimated to be about 5000 kJ/h. If the initial temperature of the room air is 10°C, determine how long it will take for the air temperature to rise to 20°C. Assume constant specific heats at room temperature. Also assume that the pressure inside the room is constant at 100 kPa.

25. Answer
Assumptions
1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical
point values of -141C and 3.77 MPa.
2 The kinetic and potential energy changes are negligible,
. 3 Constant specific heats at room temperature can be used for air. This assumption results
in negligible error in heating and air-conditioning applications.
4 The local atmospheric pressure is 100 kPa.
5 The room is air-tight so that no air leaks in and out during the process.
Properties
The gas constant of air is R = kPa.m3/kg.K (Table A-1).
Also, Cv = kJ/kg.K for air at room temperature (Table A-2).
Analysis We take the air in the room to be the system. This is a closed system since
no mass crosses the system boundary. The energy balance for this stationary constant-volume
closed system can be expressed as
Wpw
Steam
10,000 kJ/h
ROOM
4m  5m  7m
5,000 kJ/h

26. Answer or, The mass of air is Using the Cv value at room temperature,
Wpw
Steam
10,000 kJ/h
ROOM
4m  5m  7m
5,000 kJ/h
The mass of air is
Using the Cv value at room temperature,
It yields
t = 831 s

27. End of Chapter Five