T =ƒaƒy+ beƒ{^- kt (Equation 2)
where T is the temperature at time t, and a, b and k are constants with
values a = 20 „aC, b = 80 „aC and k = 2.6 ¡Ñ 10ƒ{4 sƒ{1.
Differentiate Equation 2 with respect to t and thus give values (with
appropriate units and to two significant figures) for the rate of change of
temperature with time at times of 1.0 ¡Ñ 103 s and 5.0 ¡Ñ 103 s.
To differentiate Equation 2 with respect to t, we can use the basic rules of differentiation.
Given Equation 2: T = a * y + b * e^(-kt)
First, we need to differentiate each term separately. Remember that the derivative of a constant is zero.
1. Differentiating the term "a * y" with respect to t:
Since y is a function of t, we treat it as a separate variable. The derivative of a constant times a function is the constant times the derivative of the function.
So, d/dt (a * y) = a * (dy/dt)
2. Differentiating the term "b * e^(-kt)" with respect to t:
The derivative of e^(-kt) can be found using the chain rule. If we have a function f(x) = e^(g(x)), then the derivative of f(x) is f'(x) = g'(x) * e^(g(x)).
Let g(x) = -kt, then g'(x) = -k.
So, d/dt (b * e^(-kt)) = b * (-k) * e^(-kt)
Now, let's find the derivative of Equation 2 with respect to t:
d/dt (T) = d/dt (a * y + b * e^(-kt))
= a * (dy/dt) + b * (-k) * e^(-kt)
Now, we have the expression for the rate of change of temperature with respect to time (dT/dt) in terms of the derivatives of y.
Given the values of the constants:
a = 20 °C, b = 80 °C, and k = 2.6 x 10^4 s^-1, we can use these values to calculate the rates of change of temperature at times of 1.0 x 10^3 s and 5.0 x 10^3 s.
Substituting the values into the expression for dT/dt:
dT/dt = 20 * (dy/dt) + 80 * (-2.6 x 10^4) * e^(-2.6 x 10^4 * t)
Now, plug in the values of t = 1.0 x 10^3 s and t = 5.0 x 10^3 s into the derived expression for the rate of change of temperature to find the values with appropriate units and to two significant figures.