Newton's Law of Cooling Q (1 Viewer)

[azureus88]

"a body, initially at room temperature 20% is heated so that its temperature would rise by 5degrees/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 120degrees. How long would it take to reach a temperature of 100degrees"

Im confused about the working but this is wat i got so far:

T = 20 + 100e^(-kt)

dT/dt = -k100e^(-kt) = 5

when t = 0, k = -0.05

100 = 20 + 100e^(-0.05t)
e^(-0.05t) = 4/5
-0.05t = ln(4/5)
t = 20ln(5/4) = 4.46 (2dp)

answers meant to be 32.2mins, so im way off. anyone got any brilliant ideas? thanks

 

[lyounamu]

"a body, initially at room temperature 20% is heated so that its temperature would rise by 5degrees/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 120degrees. How long would it take to reach a temperature of 100degrees"

Im confused about the working but this is wat i got so far:

T = 20 + 100e^(-kt)

dT/dt = -k100e^(-kt) = 5

when t = 0, k = -0.05

100 = 20 + 100e^(-0.05t)
e^(-0.05t) = 4/5
-0.05t = ln(4/5)
t = 20ln(5/4) = 4.46 (2dp)

answers meant to be 32.2mins, so im way off. anyone got any brilliant ideas? thanks

dT/dt = 5-k(T-20)
dT/dt = 0 when T = 120

5-k(120 - 20) = 0
5 - 100k = 0
k = 1/20
dT/dt = 5 - 1/20 ( T-20) = 6 - 0.05T

Flip dT/dt so

dt/dT = 1/(6-0.05T)
Integrate:
t + c = -1/0.05 ln(6-0.05T)

t = 0 when T=20 so c = -20 ln5
So t = 20 ln5 - 20 ln(6-0.05T)
= 20ln(5/(6-0.05T))
Then
t/20 = ln(5/(6-0.05T))
e^t/20 = 5/(6-0.05T)

Simplify that and you get:

T = 120 - 100e^-t/20
When t = 100
100 = 120 - 100e^-0.05t
100e^-0.05t = 20
e^-0.05t = 1/5
-0.05t = ln 1/5
t = 32.2

whew, there you go. Hope that it is self-explanatory.

 

[azureus88]

nice work, can u explain why dT/dt = 0 when T = 120? How does that relate to T=120 being max temperature

 

[lyounamu]

nice work, can u explain why dT/dt = 0 when T = 120? How does that relate to T=120 being max temperature

Max temp occurs when dT/dt = 0

 

[azureus88]

omg how did i miss that lol

 

[danz90]

even if i didn't know newton's law of cooling... wouldn't i still be able to get the answer using exponential growth and decay method?

 

[dolbinau]

Newton's Law of Cooling is the maths extension version of Growth/Decay I think.

 

[Mainoyi]

dT/dt = 5-k(T-20)
dT/dt = 0 when T = 120

5-k(120 - 20) = 0
5 - 100k = 0
k = 1/20
dT/dt = 5 - 1/20 ( T-20) = 6 - 0.05T

Flip dT/dt so

dt/dT = 1/(6-0.05T)
Integrate:
t + c = -1/0.05 ln(6-0.05T)

t = 0 when T=20 so c = -20 ln5
So t = 20 ln5 - 20 ln(6-0.05T)
= 20ln(5/(6-0.05T))
Then
t/20 = ln(5/(6-0.05T))
e^t/20 = 5/(6-0.05T)

Simplify that and you get:

T = 120 - 100e^-t/20
When t = 100
100 = 120 - 100e^-0.05t
100e^-0.05t = 20
e^-0.05t = 1/5
-0.05t = ln 1/5
t = 32.2

whew, there you go. Hope that it is self-explanatory.

sorry, i have some question abt this as well.
why it's (5-k)(T-20),what does 5-k mean?
im so freaks out abt this...omg..
any one can help me plz?

 

[Drongoski]

Let me attempt to explain.

You have not cited correctly; it should be 5 - k(T - 20) and not (5 - k)(T - 20)

So the (differential) equation for rate of change of temperature 'T' wrt time 't' was given by Lyounamu as: dT/dt = 5 - k(T - 20)

Now the rate of change of temperature at any time 't', viz 'dT/dt' is made up of two parts:

1) it is increasing by '5' (i.e. at 5 degrees / min)

2) it is decreasing at a rate proportional to the temperature above the room temp, i.e.
it is proportional to (T - 20); now this mean it is = k x (T-20) where k is the
constant of proportionality; the '-' in front of k reflects the decrease (as opposed to
increase) in temperature (i.e 'cooling' ... hence Newton's Law of Cooling)

Hope this helps (ming bai ma ?).

 

[dookin94]

Flip dT/dt so

dt/dT = 1/(6-0.05T)
Integrate:
t + c = -1/0.05 ln(6-0.05T)



why is it that when you integrate dt/dT you have a constant, in t + c,
but when you integrate 1/(6-0.05T) you don't have a constant?
i thought the answer should be:
t + c = -1/0.05 ln(6-0.05T) + c

and then the c's would cancel out but it didnt work and i don't know why