1. Physics 111 – Thermodynamics

2. Review Problem 1:
Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is m.

3. Solution:
W = F cos degree * S
= 45 cos 50 * 75
= 2170 Joules

4. Review Problem 2:
Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of m/s. Find the recoil velocity of the man.

5. Solution:
P=p
Mv =mv
V = - mv /m
V= - 54*25 / 88
= -1.5 m/s

6. Final Review Problem:
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second, what is the tension in the string?

7. F tension = F centripetal force = ma = m v^2/ r
Solution:
F tension = F centripetal force = ma = m v^2/ r
V= 2 pie / T = 2 pie / 5 = m/s
Ft= .25 ( 6.28 ^2 /.5) = 20 N

8. Fill in the Chart Unit: What is Constant? Value of work, W
Value of Heat exchange, Q
Value of Internal Energy Change, U
Isobaric
Isochoric
Isothermal
Adiabatic

9. Solution: Unit: What is Constant? Value of work, W
Value of Heat exchange, Q
Value of Internal Energy Change, U
Isobaric
Pressure
+/-
Isochoric
Volume
Isothermal
Temperature
Equal to Q
Equal to W
Adiabatic
Heat Transfer

10. Today ‘s Material
An ideal gas absorbs 238J of heat as it performs 845 J of work. What is the resulting change in temperature if there are 2.4 moles of an ideal gas in the system?

11. Solution: Q= +238 W=+845 U = 238 – 845 =-607 Joules U = 3/2 nRT
T = 2/3 U / nR
= 2/3 ( -607) / ( 2.4 * 8.31)
T = -20 Kelvin

12. Practice Problem 2:
A system containing an ideal gas at a constant pressure of 2.2 × 10^5 Pa gains 2431 J of heat. During the process, the internal energy of the system increases by J. What is the change in volume of the gas?

13. Solution: P = 2.2 × 10^5 Pa Q = + 2431 J U = + 1243 J
W = Q – U = =1188
W = PV
V= W/P = 1188/ 2.2 * 10 ^ 5 = 5.4 *10^ -3

14. LAST PROBLEM!!!!
What is the specific heat of a metal if its mass is 26.86g and it requires J of heat energy to raise its temperature from to 67.3 degrees Celsius .

15. 418.6 J = 26.86 (C) (67.3 – 27.4) C = .391 J/ gram Celsius
SOLUTION:
418.6 J = (C) (67.3 – 27.4) C = .391 J/ gram Celsius