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

To relate heat to temperature, we assume:

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

Our word equation is

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

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.

Using our variables and parameters, the word equation

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

becomes

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

This differential equation relates rate of change of heat and rate of change of temperature.

• The differential equation relating heat and temperature expresses a proportionality relationship between the two rates of change.
  
• We have assumed that \( c \) is independent of mass and the 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.
• If the difference in temperature between the surface of the object and the surroundings increases, then we would expect heat to be lost faster.
• Slight movement of air will facilitate a 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 the surface and its immediate surroundings.
• There is slowly moving air across the cooling object.
  

These assumptions will lead to Newton's Law of Cooling for heat exchange.

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.

Newton's law of cooling states that

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

• The \( \pm \) sign is determined by context of problem.

• Since \( Q \) denotes the rate of heat flow, we have

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

From our previous differential equation

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

Newton's law becomes

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

This can in turn be written as

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

• If coffee is 60 \( ^\circ \) C, how long will it take to cool down to 40 \( ^\circ \) C?
• If temperature of coffee is uniform throughout cup (homogeneous), then temperature will be a function of time alone (and not location as well).
• The temperature of the coffee drops because heat energy is transferred to the surrounding air, provided the surrounding air is at lower temperature.

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 condition described the first assumption is to stir the coffee. This introduces a convective effect as the heat energy is distributed. Once all the temperature inhomogeneities are removed, convection is eliminated and the coffee temperature is uniform throughout the 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, there would need to be a slight movement of air around the 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, 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 coffee cup was insulated and had a lid, then wind chill would be minimal, and difference \( \Delta U \) would be smaller due to temperature of the surroundings \( u_s \) being higher.