Ch9.2 Some Basic Physical Laws

Heat and Temperature

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  • The result of applying heat, or heat energy, to an object is to raise its temperature.
  • As an object cools, its temperature drops as it loses heat.
  • The greater the mass of the object, the more heat is required to change its temperature.

Heat and Temperature

  • For our model, need to take into account the amount of heat flowing into or out of a system.
  • Note that we observe a change in temperature due to heat flow, but we do not say that there is a temperature flow.

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Heat and Temperature

  • For our model, need to take into account the amount of heat flowing into or out of a system.
  • Thus we need to be able to relate heat and temperature, and this can be done using the specific heat of a substance.

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Relate Heat and Temperature

  • We assume the change in heat is directly proportional to the change in temperature and also the mass of the object.
  • Thus we write

\[ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, content} \end{Bmatrix} = c \times \begin{Bmatrix} mass \end{Bmatrix} \times \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, temperature} \end{Bmatrix} \]

  • The constant of proportionality \( c > 0 \) is known as the specific heat of the material.

Specific Heat

  • Let \( Q \) = rate of change of heat with time (Watts)
  • Let \( m \) = mass (kg) of material being heated or cooled
  • Let \( U \) = temperature.
  • Then the word equation

\[ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, content} \end{Bmatrix} = c \times \begin{Bmatrix} mass \end{Bmatrix} \times \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, temperature} \end{Bmatrix} \]

becomes

\[ Q = cm \frac{dU}{dt} \]

Specific Heat

  • The equation relating rate of change of heat to rate of change of temperature is

\[ Q = cm \frac{dU}{dt} \]

  • Specific heat is the constant of proportionality \( c \).
  • Specific heat is amount of heat required to raise temperature of 1kg of a substance at a given temperature by 1C.

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Table of Specific Heat Values

  • Specific heat c is not constant for large temperature range.
  • For small temperature range, can use our equation

\[ Q = cm \frac{dU}{dt} \]

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Newton's Law of Cooling

  • A mechanism for heat to be lost from an object is that of exchanging heat energy with its surroundings.
  • This takes place from the surface of the object, and thus we would expect the rate of heat loss to increase with the exposed surface area of the object, and be directly proportional to this area.
  • Further, if the difference in temperature between the surface of the object and the surroundings increases, then we would expect heat to be lost faster.

Newton's Law of Cooling

  • We therefore assume that rate of heat flow is directly proportional to temperature difference between the surface and its immediate surroundings.
  • Further, the existence of slowly moving air across the cooling object (a slight breeze) is assumed.
  • Under these conditions we have Newton's law of cooling, which works equally well with heating problems.

Newton's Law of Cooling

  • Newton's law of cooling states that

\[ \begin{Bmatrix} \mathrm{rate \, of \, heat} \\ \mathrm{exchanged \, with } \\ \mathrm{surroundings } \\ \end{Bmatrix} = \pm h S \Delta U \]

  • \( \Delta U \) = temperature difference
  • \( S \) = surface area (\( m^2 \)) from which heat is lost/gained
  • \( h \) = convective heat coefficient (Newton cooling coeff), with units Watts/\( m^2 \)/\( C \)
  • The \( \pm \) sign is determined by context of problem.

Example 9.1: Coffee Cup Problem

  • Formulate a differential equation for temperature of cup of coffee over time.
  • Compartment diagram and word equation:

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\[ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, content} \end{Bmatrix} = \begin{Bmatrix} \mathrm{rate \, heat \, lost} \\ \mathrm{to \, surroundings} \end{Bmatrix} \]

Example 9.1: Coffee Cup Problem

  • Substituting the equations of this section into word equation, we have

\[ Q = cm \frac{dU}{dt} = - h S \Delta U \]

  • This can also be written as

\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s) \]

  • Here, \( u_s \) is the temperature of surroundings.

Example 9.1: Coffee Cup Problem

  • The differential equation for our cooling coffee cup is

\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s) \]

  • Note that the solution will be decaying exponential.
  • To find the solution, can use separation of variables.
  • This analytical solution is taken up in Ch10.1.

Example 9.1: Discussion

  • Differential equation:

\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s) \]

  • We have assumed that \( h>0 \) is constant.
  • Here, \( h \) is associated with air flow around cup.
  • Rate of heat lost affected by air flow.
  • This is the wind-chill effect.

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Example 9.1: Table 9.2

  • From the table, we see that the coefficient \( h \) increases with the velocity of the air-flow passing over a metal plate.
  • This table indicates that the wind chill effect is nontrivial.

\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s) \]

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