Newton's law of cooling, $$ begin{equation} {dTover dt} = -h(T-T_s) tag {66} end{equation) SS can be...

 

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Newton's law of cooling, $$ \begin{equation} {dT\over dt} = -h(T-T_s) \tag {66} \end{equation) SS can be used to see how the temperature (T) of an object changes because of heat exchange with the surroundings, which have a temperature (T_s\). The parameter (h), with unit \(\hbox {s}^{-1}\) is an experimental constant (heat transfer coefficient) telling how efficient the heat exchange with the surroundings is. For example, (66) may model the cooling of a hot pizza taken out of the oven. The problem with applying (66) is that (h) must be measured. Suppose we have measured \(\) at \(t=0\) and \(t_1 \). We can use a rough Forward Euler approximation of (66) with one time step of length \(t_1\). SS \begin{equation*} \frac {T(t_1)-T(0)} {t_1} = -h(T(0)-T_s).\end{equation*} $$ to make the estimate $$ \begin{equation} h= \frac {T(t_1)-T(0)} {t_1(T_s-T(0))} \tp \tag{67} \end{equation} $$ a) The temperature of a pizza is 200 C at (t-0), when it is taken out of the oven, and 180 C after 50 seconds, in a room with temperature 20 C. Find an estimate of \(h) from the formula above. b) Solve (66) numerically by a method of your choice to find the evolution of the temperature of the pizza. Plot the solution. Newton's law of cooling, $$ \begin{equation} {dT\over dt} = -h(T-T_s) \tag {66} \end{equation) SS can be used to see how the temperature (T) of an object changes because of heat exchange with the surroundings, which have a temperature (T_s\). The parameter (h), with unit \(\hbox {s}^{-1}\) is an experimental constant (heat transfer coefficient) telling how efficient the heat exchange with the surroundings is. For example, (66) may model the cooling of a hot pizza taken out of the oven. The problem with applying (66) is that (h) must be measured. Suppose we have measured \(\) at \(t=0\) and \(t_1 \). We can use a rough Forward Euler approximation of (66) with one time step of length \(t_1\). SS \begin{equation*} \frac {T(t_1)-T(0)} {t_1} = -h(T(0)-T_s).\end{equation*} $$ to make the estimate $$ \begin{equation} h= \frac {T(t_1)-T(0)} {t_1(T_s-T(0))} \tp \tag{67} \end{equation} $$ a) The temperature of a pizza is 200 C at (t-0), when it is taken out of the oven, and 180 C after 50 seconds, in a room with temperature 20 C. Find an estimate of \(h) from the formula above. b) Solve (66) numerically by a method of your choice to find the evolution of the temperature of the pizza. Plot the solution.

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a Using the given values in the formula 67 we can estimate h as follows T0 200C pizza temperature at ... View the full answer