Key4Sp00

PHY 1151
Spring 2000
Fourth Hour Exam
April 26, 2000

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?

This is homework question 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 +

Te^{- 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 +

T 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

2. An ideal gas is sealed in a rigid container at 25°C and 1.0 atm. What will its temperature be when the pressure is incresed to 2.0 atm?

What will its pressure be when the temperature is increased to 50°C?

This is homework question 14.1.
The ideal gas law is
PV = nRT
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/_PoT/P = T_o/P_o
or
T = P [ ^To/_Po ] = [ ^P/_Po] T_o
T = P [ T_o/P_o ] = [ P/P_o ] T_o
Remember, these temperatures must be absolute temperatures,

T_o = 25°C = 298 K
T = [ ^P/_Po] T_o = [^{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,
T_o = 25°C = 298 K
and
T = 50°C = 323 KP = [ ^T/_To] P_o
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 have 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?

This problem is similar to the SHM and Energy Conservation lab that we did.

E = (¹/₂) k A²
E = (0.5)(8 ^N/_m) (0.10 m)²
E = 0.04 J

KE = (¹/₂) m v²
KE_max = E = (¹/₂) m v_max²
v_max² = 2 E / m
v_max² = 2 (0.04 J) /0.2 kg
v_max² = 0.4 (^J/_kg) [^{N - m}/_J] [ (kg - m/s²)/N]
v_max² = 0.4 m²/s²
v_max = 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. What frequency is heard by a stationary observer when the train approaches at 25 m/s? Use v = 340 m/s as the speed of sound in air.

This is one part of homework question 16.49 . It requires use of Equation 16.17 or 16.19. Be careful with the sign conventions; since the source is approaching, that means v_x < 0 .

f ' = (500 Hz) (1.079) = 540 Hz

Always ask if an answer is reasonable. If the frequency of the stationary train is 500 Hz and you calculate a frequency of 37 Hz or 0.73 Hz or something like that, alarms should go off! What do you expect to hear!?!?! Do you expect to hear the moving train? Of course you do! Can you hear frequencies of 37 Hz or 0.73 Hz? No!

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