Chapter 14
The Giancoli onLine Tutor
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Heat.

I. Heat Lost/Gain

Calorimetry

Heat Gained (Lost) by a substance is normally translated into an increase (decrease) in temperature of the substance. The more of the substance present the less the effect of a given amount of heat, Q. Thus Q is proportional to the product of the mass and change in temperature. The proportionality constant is known as the specific heat and denoted by "c".

Q=m*c*DT

The specific heat is normally a given. The specific heat of water is
  • 1 cal/gm*Co
  • 1 kcal/kg*Co
  • 4.16 J/gm*Co
  • 4160 J/kg*Co
    • A calorie (cal) is another unit of energy, one calorie is equal to 4.16 J. Occasionally, The heat added or removed from a substance goes into a change in phase, i.e., liquid <->gas or liquid <->solid.
    Heat of Fusion (Lf)Heat of Vaporization (Lv)
    liquid <->gas539 kcal/kg
    liquid <->solid80 kcal/kg

    Interactive Example

    As an example of how to use the above we trace the heat necessary to raise 1 kg of ice at -50 oC to steam at 150 oC. Note that the specific heat of both ice and steam is about 0.5 kcal/kg*Co. Fill in your answer to each step to the nearest kcal and click outside each time to receive feedback.
    First raise the temperature of the ice from -50 to 0 oC so that it can be melted. Q=m*c*DT Q1=
    Next we need to melt the ice. During this process the temperature does not change. Q=m*Lf Q2=
    The melted ice, now in the liquid form known as water, is at 0 oC, its temperature must now be raised to 100 oC so that it can be boiled. Q=m*c*DT Q3=
    The water at 100 oC must be turned into steam at 100 oC. Q=m*Lv Q4=
    Finally we need to raise the steam at 100 oC to steam at 150 oC. Q=m*c*DT Q5=
    TotalQ=Q1+Q2+Q3+Q4+Q5=

    Related to the above is the "science" of Calorimetry. Objects and/or fluids at different temperatures are put into contact with each other and allowed to come to a common temperature. If the system, so composed, is isolated from the outside world then any energy lost in the form of heat by the hotter constituents as they cool down will be gained by the cooler constituents as they warm up.

    You work problems using the formulas in the above table, grouping all the constituents that loose heat on the left side of the equals sign and all those that gain heat on the right side. On the left side the terms will look like

    m*c*(T1-Tf)

    where T1 is the starting temperature and Tf is the final or equilibrium temperature. On the right side the terms will look like

    m*c*(Tf-T2)

    where Tf is again the equilibrium temperature and T2 is the starting temperature.
    This is only one equation so you can only determine one thing by using it. Generally you will use it to determine Tf but Tf could be given and you could be asked to find out something else, such as, the specific heat of an unknown substance (and thereby identifying the substance).

    Interactive Example

    An unknown metal of mass 0.5 kg and temperature 450 oC is dropped into 0.88 kg of water at 0 oC, The combination comes to an equilibrium temperature of 50 oC. By consulting the table on page 421 identify the material by first finding its specific heat. Enter the name below and click outside.

    Internal Energy and Heat

    The question can be raised "How come different substances have different specific heats?" To address this question we will consider gasses. The energy in a gas is stored in its molecules and different molecules have different masses. For example, 1 kg of Helium has 8 times as many molecules as 1 kg of Oxygen but even if you had the same number of molecules they would not have the same specific heat. The reason is that when energy enters a gas via "heat", it is divided up amongst the many different ways a molecule can store the energy. The diatomic (two atoms per molecule) Oxygen molecule can store the energy in many different ways so only part of the energy goes into raising the average kinetic energy of the molecules of Oxygen and hence the temperature. The situation is much different for helium, since it is monatomic (one atom per molecule) it can only store energy in the motion of its molecules, and, hence, all the energy added to Helium goes into raising the temperature.

    The internal energy, U, for a monatomic gas easily calculated. Each molecule has an average kinetic energy of 3kT/2 so for N molecules:

    U=3NkT/2,
    using the definition of k and noting that the number of moles n of a substance is N/NA this can be rewritten as

    U=3nRT/2

    Interactive Example

    To the nearest joule what is the internal energy of 1 mol of Helium at 70 oC? Enter the answer below and click outside.

    II. Mechanisms for 'Heat Transfer'

    Heat is actually "The transfer of energy because of a temperature difference." So heat is a process for transferring energy and not a substance.

    There are three ways that this can be accomplished. Of the three, convection can not be approached quantitatively at this level so it will not be further addressed here.

    Conduction

    Conduction, by contrast , is easily analyzed at our level. Even sophisticated analysis is possible as illustrated below . Heat transfer across a barrier is proportional to the surface area, A, of the barrier and temperature difference, DT across it. It is inversely proportional to the thickness of the barrier, L. The proportionality constant, k, is called the Coefficient of Thermal Conductivity and depends on the material makeup of the barrier.
    H= DQ/Dt= k(A/L)DT/Dt
    It is useful to rewrite the above as:
    H= (A/R)DT/Dt
    where R is called the thermal resistance and equals L/k.

    The advantage of this approach is that if the barrier is constructed of various materials the formula will work by simply adding the resistances before using it.

    Interactive Activity

    This is a simulation of heat loss through a wall. It is assumed that the wall is 3 m by 10 m and 10 cm thick. You can choose the number and size of the windows, if any. You can also choose the structure of the walls as well as the inside and outside temperature.

    The purpose of this application is to illustrate the use of the formula for heat conduction.

    H= A*DT/R
    where H is the loss in watts, A is the area in m2, DT is the difference in temperature (in Celsius) between inside and outside. R is the thermal resistance, which for a single material equals L/k where k is the thermal conductivity (see table below) and L is the thickness in meters of the material. For a composite wall the thermal resistance is equal to the sum of the resistances of the individual layers. See if you can duplicate the results you find here by using your own calculations. Start with a solid wall and no windows. Next include windows and composite layers in the wall. Remember that any area occupied by windows subtracts from the area of the wall.

    wood kw= .2 w/mCo

    fiberboard kf= .06 w/mCo

    plaster board kp= .5w/mCo

    insulation ki= .03 w/mCo     		While air in a small space such as in a double pane window is a poor conductor of heat, in a large space, such as, in a wall convection current will reduce the effective resistance to zero.

    The resistance of a double pane window is approximately .6 m2Co/w.

    In a real world application one would have to include the effects of the wooden studs and other refinements.
    Number of Windows
    Size of each window in meters high by wide
    First layer of wall material
    Thickness in cms (must add to 10):
    Second layer of wall material
    Thickness in cms (must add to 10):
    Third layer of wall material
    Thickness in cms (must add to 10):
    Inside Temperature in Celsius
    Outside Temperature in Celsius
    Heat Loss in watts (J/s)

    Radiation

    All bodies in the universe radiate, most in the infrared. If a body has a temperature T (in kelvin) the energy it radiates per second, H, is given by

    H=seAT4
    where
    s   is the Stefan-Boltzmann Constant (5.67x10-8 W/m2K4)
    e   is the emissivity. It ranges in value between from ~0 for a silvered body to 1 for a black body. In most applications it will be 1.
    A is the surface area.
    If a body at temperature T1 is surrounded by an environment at temperature T2 (such as the walls of a room, or the sky) The energy exchanged is call radiant heat transfer and is given by

    H=seA(T14-T24) .
    If it is positive energy is being transferred to the body and conversely.


    Interactive Example

    On a clear night during the summer the ground temperature is 25 oC and the sky temperature is 10 oC. To the nearest megajoule how much energy is lost per 100 m2 of ground to the sky in one hour?



    © 1997-1999 Carl Adler mailto:Carl@Image-ination.com