Ch9.2 Some Basic Physical Laws

• In this section we will use physical laws to derive a relationship between the rate of change of heat and the rate of change of temperature.
  
• We will also derive Newton's law of cooling, which describes the process of heat exchange between an object and its surroundings.
  
• These two results will then combine in a differential equation that we can use for modeling a cooling coffee cup.
  

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• Our models will take into account amount of heat flowing into or out of a system.
• Applying heat to an object will raise its temperature.
  
• Similarly, its temperature will drop as it loses heat.
• How much heat is required to raise the temperature depends on mass of the object and its specific heat.

The specific heat of a material is amount of heat required to raise temperature of 1kg of a substance at a given temperature by 1C.

To relate heat to temperature, we assume:

• Change in heat is proportional to change in temperature.
• Change in heat is proportional to mass of object.

We can express the assumptions in terms of a word equation:

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

The constant \( c > 0 \) is the specific heat of material.

The variables and parameters that we will use are defined below.

• Let \( Q \) = rate of change of heat with time (Watts W = J/sec).
• Let \( m \) = mass (kg) of material being heated or cooled.
• Let \( U(t)= \) temperature (in Celcius) of material at \( t \) minutes.
• Let \( c > 0 \) be the specific heat of material.

The word equation can now be expressed as follows:

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, content} \end{Bmatrix} & = c \times \begin{Bmatrix} mass \end{Bmatrix} \times \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, temperature} \end{Bmatrix} \\ Q &= cm \frac{dU}{dt} \end{aligned} } \]

This equation relates the rate of change of heat with the rate of change of temperature.

• Our equation relating heat and temperature expresses a proportionality relationship between the two rates of change.
  
• We have assumed \( c \) is independent of mass and temperature.
  
• It has been shown that specific heat \( c \) is not constant over a large temperature range, and hence the differential equation is valid only for small temperature ranges.
  

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

• We next turn our attention to how the material loses heat.
• A mechanism for heat to be lost from an object is that of exchanging heat energy with its surroundings.
• Heat exchange takes place from the surface of the object.
• As difference in temperature between surface of object and surroundings increases, we expect heat loss to be faster.
• Slight air movement air facilitates consistent heat exchange.

The statements above for heat exchange lead to the following assumptions:

• The rate of heat loss is directly proportional to the exposed surface area of the object.
• The rate of heat flow is directly proportional to temperature difference between surface and immediate surroundings.
• There is slowly moving air across the cooling object.
  

We can express the assumptions in terms of a word equation:

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, heat} \\ \mathrm{exchanged \, with } \\ \mathrm{surroundings } \\ \end{Bmatrix} = h \times \begin{Bmatrix} \mathrm{surface \, area } \end{Bmatrix} \times \begin{Bmatrix} \mathrm{temperature \, difference} \\ \mathrm{between \, surface} \\ \mathrm{and \, surroundings } \end{Bmatrix} } \]

The constant \( h > 0 \) is the convective heat transfer coefficient, or the Newton cooling coefficient.

To obtain a formula relating heat loss with area and temperature difference, we first define the variables and parameters needed.

• Let \( S \) = surface area (\( m^2 \)) from which heat is lost/gained.
• Let \( h \) = proportionality constant = convective heat coefficient (Newton cooling coeff), with units Watts/\( m^2 \)/\( C \).
• Let \( U(t)= \) temperature (Celcius) of material at \( t \) minutes.
• Let \( u_s= \) temperature (Celcius) of surroundings, held constant.
• Let \( \Delta U = U(t) - u_s \) = temperature difference.
• Let \( Q \) = rate of change of heat with time (Watts W = J/sec).

The word equation can now be expressed as follows:

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{rate \, of \, heat} \\ \mathrm{exchanged \, with } \\ \mathrm{surroundings } \\ \end{Bmatrix} & = h \times \begin{Bmatrix} \mathrm{surface \, area } \end{Bmatrix} \times \begin{Bmatrix} \mathrm{temperature \, difference} \\ \mathrm{between \, surface} \\ \mathrm{and \, surroundings } \end{Bmatrix} \\ \\ Q &= \pm h S \Delta U \end{aligned} } \]

• The \( \pm \) sign is determined by context of problem.
• This equation relates the rate of change of heat with the surface area and temperature difference.

From the previous slide,

\[ Q = \pm h S \Delta U \]

We also know that

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

Thus Newton's law of cooling becomes

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

• Suppose we have a cup of coffee that is 60 \( ^\circ \) C.
  • How long will it take to cool down to 40 \( ^\circ \) C?
• We assume that the temperature of coffee is uniform throughout cup (homogeneous).
  • Temperature will then be a function of time alone (and not location as well).

• Determine the temperature of the coffee in the cup as a function of time.
  
• Once established, use the model to make predictions about cooling time.

We will make the following assumptions:

• The temperature of coffee is uniform throughout the cup (homogeneous).
• The temperature of the surrounding air is held constant and is lower than the temperature of the coffee.
• There is a slight movement of air about the cup.

One way to achieve the first assumption is to stir the coffee.

• This introduces a convective effect.
  
• Once temperature inhomogeneities are removed, convection is eliminated and temperature is uniform throughout cup.
  

The word equation for our coffee cup is given by

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

Using the notation of this section, our IVP has the form

\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s), \,\, \,\, U(0) = u_0 \]

• The analytical solution can be found using separation of variables, as will be shown in Ch10.1.
• We can observe from form of IVP and from previous models that solution will be decaying exponential.

• This model would be reasonable in a room with a controlled environment.
• That is, the air temperature is controlled by a thermostat, and there is enough air volume for the ambient temperature to be unaffected by the heat emanating from the cup.
• Also, we assume slight air movement around cup, which is reasonable since most rooms do have some air flow.
• If there was an open window with a breeze, or a door that frequently allows a breeze to enter the room, then there may be more of a wind chill effect on the cup.
  

• From the table below, 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.
  
• If the coffee cup was insulated and had a lid, then the wind chill would be minimal, and the difference \( \Delta U \) could be considered smaller if \( u_s \) was measured inside the cup, perhaps on the underside of the lid.