PHY 1151
Summer 2003
Fourth Hour Exam
July 9, 2003
Statistics:
High: 98
Mean: 84
Low: 21
1. A roasted turkey cools from 85°C to 80°C in 10 min when sitting in a 25°C room. How long does it require to cool from 85°C to 55°C?
Remember homework pblm 13.30 ?
This requires an application of Newton’s Law of Cooling, equation 13.3,
T(t) = T_{sur} + T
e^{–t/}
T is the initial
temperature difference of the turkey and its surroundings;
T = 85°C – 25°C = 60 C°
Knowing that it cools from 85°C to 80°C allows us to solve for the “time constant” ^{} in this equation;
T(t) = T_{sur} + T
e ^{– t/}
T(10 min) = 80°C = 25°C + (60 C°) e ^{– (10 min)/}
55°C = (60 C°) e ^{– (10 min)/}
^{55°C}/_{60°C} = e^{ – (10 min)/}
^{55}/_{60} = e^{ – (10 min)/}
0.9167 = e ^{– (10 min)/}
e ^{– (10 min)/}
= 0.9167
ln [ e ^{– (10 min) / }
] = ln [0.9167]
- (10 min) / ^{}
= - 0.0870
10 min / ^{}
= 0.0870
^{} = (10 min)/0.0870
^{} = 115
min
Now we know the time constant ^{} and we can use Newton’s Law of Cooling to go back and solve for t, the time, when T, the temperature, is 55°C.
T(t) = T_{sur} +DT
e^{–t / }
T(t) = 55°C = 25°C + (60 C°) e ^{– t / (115 min)}
55°C = 25°C + (60 C°) e ^{– t / (115 min)}
30°C = (60 C°) e ^{– t / (115 min)}
^{30°C}/_{60°C} = e^{ – t / (115 min)}
^{30}/_{60} = e^{ – t / (115 min)}
0.50 = e^{ – t / (115 min)}
ln [ 0.50 ] = ln [ e ^{– t / (115 min) }]
–0.693 = – t / (115 min)
0.693 = t / (115 min)
t = (0.693)(115 min)
t = 78 min
b) What will its pressure be when the temperature is increased to 50°C ?
Warning: Be sure you use the right temperature scale!
The ideal gas law is
P V = n R T
Being in a rigid container, the gas’ volume does not change; V = V_{o}= constant. That means we can use the ideal gas law as
^{T}/_{P} = ^{To}/_{Po}
_{ or T = P [ To/Po ] = [ P/Po ] To Remember, these temperatures must be absolute temperatures, To = 25°C = (25 + 273) K = 298 K (Remember, we discussed in class that the 302 K I had on the web was incorrect). T = [ P/Po ] To = [2.0 atm/1.0 atm] [298 K] = [ 2 ] [ 298 K] = 596 K T = 596 K = (596 – 273)°C = 323°C = T Or, while V = const, we can write the ideal gas law as P/T = Po/To or P = T [ Po/To ] = [ T/To ] Po Remember, these temperatures must be absolute temperatures, To = 25°C = 298 K and T = 50°C = (50 + 273) K = 323 K P = [ T/To ] Po P = [ 323 K/298 K ] (1 atm) P = 1.08 atm 3 .. An air track glider has a mass of m = 0.200 kg and is attached to spring which has an effective spring constant of k = 8.0 N/m. It oscillates with an amplitude of A = 0.10 m. a) What is the total energy of this simple harmonic oscillator? b) What is the speed of the glider as it passes through equilibrium? c) What is the period of this simple harmonic oscillator? E = (1/2) k A2 E = (0.5)(8 N/m) (0.10 m)2 E = 0.04 J KE = (1/2) m v2 KEmax = E = (1/2) m vmax2 vmax2 = 2 E / m vmax2 = 2 (0.04 J) /0.2 kg vmax2 = 0.4 (J/kg) [N-m/J] [ (kg-m/s2)/N] vmax2 = 0.4 m2/s2 vmax = 0.632 m/s T = 2 SQRT{ m/k ] T = 2 SQRT{ 0.2 kg/(8 N/m) ] T = 2 SQRT{ 0.025 s2 ] T = 2 (0.16 s) T = 0.99 s 4. A train whistle sounds at 500 Hz (when stationary, etc). The speed of sound in still air is 340 m/s. i) What frequency is heard by a stationary observer when the train approaches at 25 m/s? If the train approaches, you know you will hear a HIGHER frequency. ii) What frequency is heard by a stationary observer when the train leaves (or moves away) at 25 m/s? If the train leaves, you know you will hear a LOWER frequency. f ’ = f [(v + vobs)/(v + vsource)] f ’ = (500 Hz) [(340 m/s + 0) / (340 m/s – 25 m/s)] vsource < 0 for approach vsource > 0 for moving away Remember, for approach f ’ < fo f ’ = (500 Hz) (340/315) f ’ = (500 Hz)(1.079) f ’ = 540 Hz f’ = f [(v + vobs)/(v + vsource)] f’ = (500 Hz) [(340 m/s + 0) / (340 m/s + 25 m/s)] vsource < 0 for approach vsource > 0 for moving away Remember, for leaving, f ’ > fo f ’ = (500 Hz) (340/365) f ’ = (500 Hz)( 0.931 ) f ’ = 466 Hz Concept questions: i. What is heat? Heat is energy that is transferred because of a difference in temperature. ii. What is the efficiency of a heat engine? That is, define “efficiency” in words. Efficiency is the ratio of the output work divided by the heat absorbed from the high temperature reservoir. iii. We know metal expands when it is heated. If a hole is drilled in a metal plate and the plate is then heated, what happens to the size of the hole? The hole also expands. iv. What does ultrasonic mean? Ultrasonic means having a frequency greater than that of human hearing (about 20 kHz). v. What is resonance? A resonance occurs when a system responds with a large amplitude when it is driven or excited at a particular frequency, usually called its “natural frequency”. | Return to Calendar | }