PHY 1151
Summer 2003
Fourth Hour Exam
July 9, 2003
Statistics:
High:
Mean:
Low:
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 Newtons Law of Cooling, equation 13.3,
T(t) = Tsur + T
et/
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) = Tsur + 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 Newtons Law of Cooling to go back and solve for t, the time, when T, the temperature, is 55°C.
T(t) = Tsur +DT
et / b) What will its pressure be when the temperature is increased to 50°C ? P V = n R T Being in a rigid container, the gas volume does not change; V
= Vo= 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(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
2. An ideal gas is sealed in a rigid container at 25°C and 1.0 atm.
That is, its volume will remain constant.
a) What will its temperature be when the pressure is increased to 2.0 atm ?
Warning: Be sure you use the right temperature scale!
The ideal gas law is
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
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 = 1.12 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