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Solution for Lab 3-1

Virtual lab on chapters 13 and 14

Note: In each of the following, I am looking for main ideas only - not interested in whether you got a "right" number or not.

 

For all virtual labs, submit a written report including the following specific paragraph headings

1. Requirement, 2. Observations, 3. Principle, etc, being demonstrated, 4. Calculations, 5. Answer

 

Chapter 14: Heat: Physlets Chapter 19

 

Problem 19.1: Specific Heat, Work, Energy

 

This is an example of the mechanical equivalent of heat. Joule determined that a given amount of work was equivalent to a particular amount of heat input.

 

As the 100-kg red mass drops, a paddle turns in a liquid and the liquid heats up (position is given in meters and time is given in seconds).  The dimension of the container that holds the liquid that you cannot see (into the screen) is 0.1 m.  The density of the liquid is 920 kg/m3.

Joule used a version of this device to determine the equivalence between heat and work.

 

What is the heat capacity of the liquid in the animation?

 

Basic Approach

1.  Calculate the volume

2.  Using the calculated volume and specified density, find the mass of the liquid

3.. Find the Mechanical energy of the dropping mass (all potential energy) which is given by mgh - it will be in Joules

4.  Find delta T (final temperature - initial) in C

5.  Convert temperature delta to K

6. Use the following equation

 

Q = cmDT ð  c = Q/mDT     where Q is the mechanical energy found above coverted to heat energy

 

  

 

 

Problem 19.3: Thermal Expansion

Note that the rod expands in both directions and you are measuring for one end - must double this.

 

A rod is resting on a surface (you see the top view) and is attached to the surface at its middle (position is given in meters and time is given in minutes).  The animation shows you both the rod and a magnified view of the right end.  What is the coefficient of linear expansion of the rod?

 

Basic Approach

1.  Note tht the rod expands in both directions, so the delta L observed (for one end only) must be doubled.

2.  Use the expression for linear expansion

      DL = aL0DT 

3. Make various observations and observe the delta L, initial L, and delta T. From this, calculate alpha. You should get the same value

     for different observations

 

 

  

 

Problem 19.6: Specific Heat and Latent Heat of Fusion

Note: For t = 3.1 to 12.6, T = 597 K.

From t = 0 to t = 3.1, delta T = 57

 

This animation shows a container holding 1 kg of a solid heated by a heater that is delivering 2400 watts to the material (temperature is given in kelvin and time is given in seconds).  Find the specific heat capacity of the material and its latent heat of fusion.  Ignore the heat capacity of the container.

 

1. For the constant T portion, T = 597 K (given)

2. Power = Q/t (chapter 4) ð Q = Pt for above time period = (2400)(9.5) = 22,800 J

3. Q = mL ð LF = Q/m = 22,800 J/1 kg = 22.8 kJ/kg

4. From t = 0 to 3.1, DT = 57 (given)      Q = Pt = (2400)3.1) = 7,440 J

5. Q = mcDT ð c = Q/mDT = 7,440/(1)(57) = 130 J/kg K 

     

  

 

Chapter 13: Temperature and Kinetic Theory: Physlets Chapter 20

 

Problem 20.5: Ideal Gas Law Analysis

Note: Observe the changes in P and V

 

If the gas inside the container being compressed is an ideal gas, what is wrong with this animation?  The pressure is shown on the gauge (in atmospheres) and is caused by the black block sitting on the piston (position is given in centimeters). 

 

Basic Approach

1.  PV = nRT for an ideal gas

2.  If Pressure doubles and V is divided by 2, then T should reman constant - it does not