No problem, happy to help! Good luck with your project.

[sunmoonlight]

Homework Statement

Find the value of k

Relevant Equations

T(t) = TA + (To - TA) e ^-kt

What's the value of k, given T(t) = 51 degrees, TA = 19 degrees, To = 80 degrees and t = 1200 s?
Should the value of k be positive or negative if an object is cooling down?

 

[renormalize]

Homework Statement: Find the value of k
Relevant Equations: T(t) = TA + (To - TA) e ^-kt

What's the value of k, given T(t) = 51 degrees, TA = 19 degrees, To = 80 degrees and t = 1200 s?
Should the value of k be positive or negative if an object is cooling down?

What must the sign of $k$ be in order that the object cools as time increases, i.e., so that $T(t_{2})<T(t_{1})$ for $t_{2}>t_{1}$?

 

[sunmoonlight]

k should be positive, but when I plotted a graph of ln (T(t) - TA) vs t, I got a graph of negative k

 

[haruspex]

k should be positive, but when I plotted a graph of ln (T(t) - TA) vs t, I got a graph of negative k

Do you mean you got a graph of negative slope? That slope is not k.
Rearrange the equation into the form kt=… to find out what you should be plotting.

 

[renormalize]

k should be positive, but when I plotted a graph of ln (T(t) - TA) vs t, I got a graph of negative k

Why not just solve the logarithmic-form of the cooling law $kt=\ln\left(\frac{T\left(0\right)-T_{A}}{T(t)-T_{A}}\right)$ for k directly? (In this form it's clear that $k$ must be positive because the argument of the log is greater than 1.)

 

[sunmoonlight]

Do you mean you got a graph of negative slope? That slope is not k.
Rearrange the equation into the form kt=… to find out what you should be plotting.

According to this source, the gradient is k (which is a negative value)

 

[malawi_glenn]

According to this source, the gradient is k (which is a negative value)

Note the logarithmic vertical axis

 

[renormalize]

According to this source, the gradient is k (which is a negative value)

Where is $k$ mentioned on that screenshot? As implied in post #45, $k$ is proportional to the negative of $\ln\left(T(t)-T_{A}\right)$.

 

[haruspex]

According to this source

And what does the label on the y axis say?

 

[sunmoonlight]

ln T(t) - TA

And what does the label on the y axis say?

 

[jbriggs444]

ln T(t) - TA

Which is not $k$.

Now, how do you get $k$ from $\ln (T(t) - T_A)$?

Note that it is normally incorrect to take the logarithm of a quantity which has units. It is only proper to take the logarithm of a pure number such as a ratio. However $\ln (\frac{a}{b}) = \ln(a) - \ln(b)$, taking the log of a quantity with units simply has the effect of shifting the graph up or down by the undefined quantity: $\ln\ \text{<unit>}$.

It is also "challenging" to take the log of a negative number. So rather than simplifying as:$$\ln (\frac{-3}{-2}) = \ln (-3) - \ln(-2)$$one would be better served to simplify to $$\ln (\frac{-3}{-2}) = \ln (\frac{3}{2}) = \ln(3) - \ln(2)$$This will handle the case where $T(t)$ is less than ambient.

 

[kuruman]

Should the value of k be positive or negative if an object is cooling down?

To me, this question is meaningless and could be confusing because it is the argument of the exponential that counts. Parameter $k$ is the inverse of a time constant and, as such, it should be positive. With $k$ positive and $t$ also positive, the argument of the exponential must be negative when the object is cooling down and positive when it is heating up. In other words, the negative sign in front of the argument $kt$ is or is not there depending on what one wishes to describe.

 

[jbriggs444]

To me, this question is meaningless and could be confusing because it is the argument of the exponential that counts. Parameter $k$ is the inverse of a time constant and, as such, it should be positive.

Similar to how $g$ is positive and when up is positive, free fall has $a=-g$.

 

[haruspex]

According to this source, the gradient is k (which is a negative value)

$T(t) = T_A + (T_0 - T_A) e ^{-kt}$
$T(t) -T_A = (T_0 - T_A) e ^{-kt}$
$\ln(T(t) -T_A) =\ln( T_0 - T_A)+(-kt)$
So what will be the gradient of $\ln(T(t) -T_A)$ plotted against $t$?

 

[sunmoonlight]

$T(t) = T_A + (T_0 - T_A) e ^{-kt}$
$T(t) -T_A = (T_0 - T_A) e ^{-kt}$
$\ln(T(t) -T_A) =\ln( T_0 - T_A)+(-kt)$
So what will be the gradient of $\ln(T(t) -T_A)$ plotted against $t$?

-k

 

[sunmoonlight]

To me, this question is meaningless and could be confusing because it is the argument of the exponential that counts. Parameter $k$ is the inverse of a time constant and, as such, it should be positive. With $k$ positive and $t$ also positive, the argument of the exponential must be negative when the object is cooling down and positive when it is heating up. In other words, the negative sign in front of the argument $kt$ is or is not there depending on what one wishes to describe.

I got that. Thanks a lot.