A cup of coffee cools according to Newton's law of cooling (3) in section 1.3. $\frac{dT}{dt} = k(T - T_m)$

Using data from the temperature graph $T(t)$ provided in the figure, we need to estimate the constants $T_m$ , $T_0$ , and $k$ in the model of the form: $\frac{dT}{dt} = k(T - T_m), \quad T(0) = T_0$

Question :

A cup of coffee cools according to newton's law of cooling (3) in section 1.3. $\frac{dt}{dt} = k(t - t_m)$

using data from the temperature graph $t(t)$ provided in the figure, we need to estimate the constants $t_m$ , $t_0$ , and $k$ in the model of the form: $\frac{dt}{dt} = k(t - t_m), \quad t(0) = t_0$

A cup of coffee cools according to newton's law of cooling (3) in section 1.3. $ | Doubtlet.com

Solution:

Neetesh Kumar | November 08, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh1Sec03 (Homework) Question - 2
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Step-by-Step-Solution:

$T$ is the temperature of the coffee,
$T_m$ is the ambient temperature,
$k$ is a constant that depends on the characteristics of the coffee and the environment.

Step 1: Determine Constants from the Graph

Estimate $T_m$ :
- $T_m$ represents the ambient temperature. From the graph, the temperature $T$ approaches a constant value as $t$ increases. This value is $T_m$ .
- For example, if the graph stabilizes around $75^\circ C$ after sufficient time, we estimate: $T_m = 75$
Estimate $T_0$ :
- $T_0$ is the initial temperature of the coffee at $t = 0$ . From the graph, find the value of $T$ at $t = 0$ .
- For instance, if $T(0)$ is around $175^\circ C$ : $T_0 = 175$
Estimate $k$ :
- To estimate $k$ , we need to use the cooling data. Select two points from the graph to find the slope of $T(t)$ at a specific time.
- If we have a point $(t_1, T_1)$ and another point $(t_2, T_2)$ , we can calculate the average rate of change of temperature over that time interval: $k \approx -\frac{1}{T_1 - T_m} \cdot \frac{T_2 - T_1}{t_2 - t_1} = -0.07$

Conclusion

Using the estimates obtained from the graph:

$T_m = 75$
$T_0 = 175$
$k = -0.07$

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