Rate of change question (1 Viewer)

[Menomaths]

Newtons law of cooling states that the rate of cooling of a body is proportional to the excess of its temperature above its surroundings i.e dT/dt = -kt where T is the excess of temperature above its surroundings at any time t. The temperature of a room is 25 and the outdoor temperature is 5. A thermometer which has been kept in the room is moved outside. In five minutes the thermometer reading is 15. What would its reading be in another 5 minutes?

Such confusing questions...

 

[anthony_wheels]

I'm pretty sure Newton's rate of cooling is in three unit... I haven't learnt it yet so I dunno how to do it :/

 

[Menomaths]

It has nothing to do with his law the information needed is given, I'm just to blind to see it

 

[anthony_wheels]

Ahk, fair enough... I can't either haha

 

[bedpotato]

dT/dt = -kT
T = Ae^(-kt)

Can you work it out from there?

 

[Hypem]

Is the answer 5?

dT/dt = -kT
T = Ae^(-kt)

Can you work it out from there?

The question says

dT/dt = -kt

vs.

dT/dt = -kT

It appears your one makes sense as the excess T would decrease so the rate of change would decrease with it?

 

[bedpotato]

The question says

dT/dt = -kt

vs.

dT/dt = -kT
It appears your one makes sense as the excess T would decrease so the rate of change would decrease with it?

Based on that it'd be 9

Sorry, that was a mistake...
Thanks for pointing it out.


"T is the excess of temperature above its surroundings at any time t."

If the outside temperature if 5, and the reading on the thermometer is 25, then the excess of temperature would be 20.
After 5 minutes, the outside temperaure is still 5, but the reading on the thermometer is 15. So the excess of temperature would be 10.

So, at t = 0, T = 20. Therefore, A = 20
At t = 5, T = 10.

 

[Hypem]

Sorry, that was a mistake...
Thanks for pointing it out.

Differentiating Ae^-kt gives you -k(Ae^-kt) = -kT

....

 

[Hypem]

Sorry, that was a mistake...
Thanks for pointing it out.


"T is the excess of temperature above its surroundings at any time t."

If the outside temperature if 5, and the reading on the thermometer is 25, then the excess of temperature would be 20.
After 5 minutes, the outside temperaure is still 5, but the reading on the thermometer is 15. So the excess of temperature would be 10.

So, at t = 0, T = 20. Therefore, A = 20
At t = 5, T = 10.

Oh yeah, so it would be 5 then...

 

[bedpotato]

Differentiating Ae^-kt gives you -k(Ae^-kt) = -kT

....

Yeah. There's a mistake in the question then.

It should be dT/dt = -kT

 

[Menomaths]

Answer is 10, how'd you work that out
So blank at the moment

 

[bedpotato]

After 10 minutes, T=5, and the outside temperature is 5. Since T is the excess of temperature, the reading on the thermometer is 5 + 5 = 10.

Therefore, after another 5 minutes, the reading on the thermometer is 10.

 

[Menomaths]

Ahh, thanks