Physical Application of Calculus Q (2 Viewers)

[lyounamu]

Q 4 on pg 129, Fitzpatrick 3 Unit book.

I am struggling how to get the answer.

This is the question:

A body, initially at room temperature 20 degrees Celsius, is heated so that its temperature would rise by 5 degrees Celsius/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120 degrees Celsius. How long would it take to reach a temperature of 100 degrees Celsius?

Thanks in advance.

 

[vds700]

Q 4 on pg 129, Fitzpatrick 3 Unit book.

I am struggling how to get the answer.

This is the question:

A body, initially at room temperature 20 degrees Celsius, is heated so that its temperature would rise by 5 degrees Celsius/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120 degrees Celsius. How long would it take to reach a temperature of 100 degrees Celsius?

Thanks in advance.

hey namu here u go

dT/dt = 5 - k(T - M) (5 is the heating and the k(T-M) is the cooling, M is the surrounding temp)
..... you invert this and integrate with respect to T and i got
T = (1/k)[5 + kM - e^-kt(5 - 20k + kM)]
as t -> infinity, T->120, therefore the e^-kt part -> zero. Also M = 20 as the room temp is 20.

120 = (1/k)(5 + 20k)
120k = 5 + 20k
100k = 5, k = 1/20

Putting this in the original expression for T you get
T = 20[6 - 5e^(-t/20)] ... when T = 100
100 = 20[6 - 5e^(-t/20)]
5 = 6 - 5e^(-t/20)
5e^(-t/20) = 1
e^(-t/20) = 0.2
-t/20 = ln(0.2)
therefore t = 32.2 minutes

 

[lyounamu]

hey namu here u go

dT/dt = 5 - k(T - M) (5 is the heating and the k(T-M) is the cooling, M is the surrounding temp)
..... you invert this and integrate with respect to T and i got
T = (1/k)[5 + kM - e^-kt(5 - 20k + kM)]
as t -> infinity, T->120, therefore the e^-kt part -> zero. Also M = 20 as the room temp is 20.

120 = (1/k)(5 + 20k)
120k = 5 + 20k
100k = 5, k = 1/20

Putting this in the original expression for T you get
T = 20[6 - 5e^(-t/20)] ... when T = 100
100 = 20[6 - 5e^(-t/20)]
5 = 6 - 5e^(-t/20)
5e^(-t/20) = 1
e^(-t/20) = 0.2
-t/20 = ln(0.2)
therefore t = 32.2 minutes

Thank you so much! I got the equation for dT/dt but I could not continue from there. Thanks again.

By the way, how are you going with your studies?

EDIT: Do you know how to do Q7 on the same page? Sorry.

 

[Iruka]

You can find k a little easier by realising that when T=120, dT/dt = 0.