Tuesday, May 16, 2017

Solution to Lab Exercise...051717

SOLVED PROBLEMS #1
1. When 2,100 J is removed from 0.10 kg of a substance, its temperature is observed
    to decrease from 40oC to 10oC.  What is the specific heat of the substance?

    Solution:  Using the formula: Q  =  mcΔT where Q =  2,100 J

2,100 J  =  0.10 kg (c) (40 – 10)oC

c  =  700 J/kg-oC

2. How much heat is required to raise an 800 gm copper pan from 15oC to 90oC if the
    pan contains 1.0 kg of water?

    Solution:  Since both the pan and the water absorb heat, the total heat required is:

Q =  Heat absorbed by the pan  +  Heat absorbed by the water

   =  0.8 kg (386 J/kg-K) (90 – 15)  +  1.0 kg (4185 J/kg-K) (90 – 15)
   =  337,000J

3. In a calorimetry experiment, 0.50 kg of a metal at 100oC is added to 0.50 L of
    water at room temperature in an aluminum calorimeter cup.  The cup has a mass of
    250 gm.  If the final temperature of the mixture is 25oC, what is the specific heat of
    the metal?
    Solution:

In this problem, the calorimer and the water in it will absorb the heat released
by the hot  metal.

0.50 kg (c) (100 – 25)oC = 0.5 kg (4185) (25 – 20) + 0.250 (920) (25 – 20)

c  =  309.67 J/ kg-oC

4. Determine the amount of heat a freezer has to remove from 1.5 kg of water at 22oC                  
    to make ice at – 10oC.

    Solution:
a) Heat required to change the temperature of water at 22oC to 0oC

Q1  =  mcΔT  =  (1.5 kg) (4186 J/kg-oC)(22 – 0)oC
     =  138.14 J

b) Heat required to change water at 0oC to ice at 0oC

Q2  =  mLf  =  (1.5 kg)(3.33 x 105 J/kg)  
                                   =  4.995 x 105 J

c) Heat to be removed to lower ice from 0oC to – 10oC

Q3  = mcΔT  =  (1.5 kg) (2100 J/kg-oC) [0 – (–10)oC]
      = 3.15 x 104 J

Total Heat required:  Qtotal  =  138.14 J +  4.995 x 105 J  +  3.15 x 104 J  

Qtotal  =  5.31 x 105 J

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SOLVED PROBLEMS #1
1. When 2,100 J is removed from 0.10 kg of a substance, its temperature is observed
    to decrease from 40oC to 10oC.  What is the specific heat of the substance?

    Solution:  Using the formula: Q  =  mcΔT where Q =  2,100 J

2,100 J  =  0.10 kg (c) (40 – 10)oC

c  =  700 J/kg-oC

2. How much heat is required to raise an 800 gm copper pan from 15oC to 90oC if the
    pan contains 1.0 kg of water?

    Solution:  Since both the pan and the water absorb heat, the total heat required is:

Q =  Heat absorbed by the pan  +  Heat absorbed by the water

   =  0.8 kg (386 J/kg-K) (90 – 15)  +  1.0 kg (4185 J/kg-K) (90 – 15)
   =  337,000J

3. In a calorimetry experiment, 0.50 kg of a metal at 100oC is added to 0.50 L of
    water at room temperature in an aluminum calorimeter cup.  The cup has a mass of
    250 gm.  If the final temperature of the mixture is 25oC, what is the specific heat of
    the metal?
    Solution:

In this problem, the calorimer and the water in it will absorb the heat released
by the hot  metal.

0.50 kg (c) (100 – 25)oC = 0.5 kg (4185) (25 – 20) + 0.250 (920) (25 – 20)

c  =  309.67 J/ kg-oC

4. Determine the amount of heat a freezer has to remove from 1.5 kg of water at 22oC                  
    to make ice at – 10oC.

    Solution:
a) Heat required to change the temperature of water at 22oC to 0oC

Q1  =  mcΔT  =  (1.5 kg) (4186 J/kg-oC)(22 – 0)oC
     =  138.14 J

b) Heat required to change water at 0oC to ice at 0oC

Q2  =  mLf  =  (1.5 kg)(3.33 x 105 J/kg)  
                                   =  4.995 x 105 J

c) Heat to be removed to lower ice from 0oC to – 10oC

Q3  = mcΔT  =  (1.5 kg) (2100 J/kg-oC) [0 – (–10)oC]
      = 3.15 x 104 J

Total Heat required:  Qtotal  =  138.14 J +  4.995 x 105 J  +  3.15 x 104 J  

Qtotal  =  5.31 x 105 J  

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Friday, May 15, 2015

Homework , fluids/heat

fluids



7. A 5/8 inch diameter garden hose is used to fill a round swimming pool 6.1 m in
    diameter.  How long will it take to fill the pool to a depth of 1.2 m is water issues
    from the hose at a speed of 0.4 m/sec?









8. A very large vat of water has a hole 1 cm in diameter located at a distance 1.8 m
    below the water level.  How fast does water exit the hose?












9. A house with its own well has a pump in the basement with an output pipe of
    inner radius of 6.3 mm.  Assume that the pump can maintain a gauge pressure of
    410 kPa in the output pipe.  A shower head in the second floor (6.7 m above the
    pump’s output pipe) has 36 holes, each of radius 0.33 mm.  The shower is on “full
    blast” and no other faucet in the house is open.  a) Ignoring viscosity, with what
    speed does water leave the showerhead?  b) With what speed does water move
    through the output of the pump?









10. A water tower supplies water through the plumbing in a house.  A 2.54 cm
    diameter faucet in the house can fill a cylindrical container with a diameter of
    44 cm and a height of 52 cm in 25 sec.  How high above the faucet is the top of the
    tower?  Assume that the diameter of the tower is so large compared to that of the
    faucet that the water at the top of the tower does not move.


heat

7. A hollow aluminum cylinder has an internal capacity of 2.00 liters at 20oC. It is
    completely filled with turpentine and then slowly warmed to 80oC.  The average 
    coefficient of linear expansion of aluminum is 24 x 10–6/oC and the average
    volume expansion coefficient of turpentine is 9 x 10–4/oC)  a) How much 
    turpentine overflows?  b) If the cylinder is then cooled back to 20oC, will its
    volume be the same?

   






8. What temperature change would cause a 0.10% increase in the volume of a
    quantity of water (β = 2.1 x 10–4/oC)  that was initially at room temperature?            












9. A thin spherical shell of silver has an inner radius of 2 x 19–2 m when the
    temperature is 18oC.  The shell is heated to 147oC.  Find the change in the interior
    volume of the shell. The coefficient of volume expansion of shell is 57 x 10–6/oC.












10. A brass rod and an aluminum rod of the same diameter are each separately
    attached to an immovable wall opposite each other.  The lengths of the rods are
    2.0 m and 1.0 m, respectively.  When the temperature is 28oC, the air gap between
    the rods is 1.3 x 10–3  m.  At what temperature will the gap be closed?
 

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Saturday, April 13, 2013

Summner 2015 Physics 101 Handouts

Phys 1 - College Physics
Summer 2015

for BSIT/Psyc/Pharma

Click on the link below to download the handouts.

Physics 101AH Lesson1 Handouts

Physics 101AH Lesson2 Handouts

Physics 101AH Lesson 3a Handouts

Physics 101AH Lesson 3b Handouts

Physics 101AH Lesson 4 Handouts

Physics 101AH Lesson 5 Handouts

Midterm Handouts


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Wednesday, January 9, 2013

Physics 101AH: Lesson 5

Physics 101AH: Lesson 5
Work, Power and Energy
Click here to download document


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Monday, January 7, 2013

Physics 101AH: Lesson 4

Physics 101AH: Lesson 4
Forces
Click here to download the document


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Saturday, December 15, 2012

Physics 101 Sample Problems with Solutions - Kinematics


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  1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
    Given:
    a = +3.2 m/s2
    t = 32.8 s
    vi = 0 m/s
    Find:
    d = ??
    d = vi*t + 0.5*a*t2
    d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2
    d = 1720 m
  2. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
    Given:
    vi = 18.5 m/s
    vf = 46.1 m/s
    t = 2.47 s
    Find:
    d = ??
    a = ??
    a = (Delta v)/t
    a = (46.1 m/s - 18.5 m/s)/(2.47 s)
    a = 11.2 m/s2
    d = vi*t + 0.5*a*t2
    d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2
    d = 45.7 m + 34.1 m
    d = 79.8 m
    (Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)
  3. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
    Given:
    vi = 0 m/s
    d = -1.40 m
    a = -1.67 m/s2
    Find:
    t = ??
    d = vi*t + 0.5*a*t2
    -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2
    -1.40 m = 0+ (-0.835 m/s2)*(t)2
    (-1.40 m)/(-0.835 m/s2) = t2
    1.68 s2 = t2
    t = 1.29 s
  4. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
    Given:
    a = -9.8 m/s2
    vf = 0 m/s
    d = 2.62 m
    Find:
    vi = ??
    vf2 = vi2 + 2*a*d
    (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)
    0 m2/s2 = vi2 - 51.35 m2/s2
    51.35 m2/s2 = vi2
    vi = 7.17 m/s
  5. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
    Given:
    a = -9.8 m/s2
    vf = 0 m/s
    d = 1.29 m
    Find:
    vi = ??
    t = ??
    vf2 = vi2 + 2*a*d
    (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)
    0 m2/s2 = vi2 - 25.28 m2/s2
    25.28 m2/s2 = vi2
    vi = 5.03 m/s
    To find hang time, find the time to the peak and then double it.
    vf = vi + a*t
    0 m/s = 5.03 m/s + (-9.8 m/s2)*tup
    -5.03 m/s = (-9.8 m/s2)*tup
    (-5.03 m/s)/(-9.8 m/s2) = tup
    tup = 0.513 s
    hang time = 1.03 s
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