Newton's Law of Cooling Question

[Skelly4444]

A cup of liquid at 5 degrees Celsius is removed from a fridge and placed in a room which is at 19 degrees Celsius.
The initial rate of cooling is 5.7 degrees Celsius per minute.

(a) Show that dx/dt = 0.3(19-x)

(b) Find the time taken for the liquid to reach the ambient temperature.

I'm really struggling to agree with the book on this one. Struggling with part (a) to be honest.

I separated the variables, integrated both sides and arrived at a value of -ln|14| for the constant of integration.

The book's answer states approximately 11 minutes for part (b) and I cannot get close to that answer, I obtain a strange value of -1.018.

Any guidance would be much appreciated.

 

[Deleted member 4993]

A cup of liquid at 5 degrees Celsius is removed from a fridge and placed in a room which is at 19 degrees Celsius.
The initial rate of cooling is 5.7 degrees Celsius per minute.

(a) Show that dx/dt = 0.3(19-x)

(b) Find the time taken for the liquid to reach the ambient temperature.

I'm really struggling to agree with the book on this one. Struggling with part (a) to be honest.

I separated the variables, integrated both sides and arrived at a value of -ln|14| for the constant of integration.

The book's answer states approximately 11 minutes for part (b) and I cannot get close to that answer, I obtain a strange value of -1.018.

Any guidance would be much appreciated.

To use Newton 's law, did you start from the DE or the solution of the DE ?

Please share your "starting" equation, defining all the variables and assumptions.

 

[Skelly4444]

I started with the equation dx/dt=0.3(19-x) and separated the variables from here. I cannot see where the initial rate of cooling fits into it all?
I can see that dividing 5.7 by 19 gives us the 0.3 value.

To be honest, I'm not entirely sure how to answer any of this question. I need someone to go through it all.

 

[Deleted member 4993]

I started with the equation dx/dt=0.3(19-x) and separated the variables from here. I cannot see where the initial rate of cooling fits into it all?
I can see that dividing 5.7 by 19 gives us the 0.3 value.

To be honest, I'm not entirely sure how to answer any of this question. I need someone to go through it all.

The constant of solution is determined from that (initial rate)

Please share your work for part (a).

 

[Skelly4444]

After separating the variables I get the following:

-ln(19-x) = 0.3t + C

When t= 0 , x=5

Therefore -ln(14) = C

This gives us the equation below which is the correct equation according to the book.

x = 19 - 14e^-0.3t

Part (b) asks us to find to the nearest degree, the time taken for the liquid to reach ambient temperature.

I cannot see how they arrive at an answer of approximately 11 minutes?

 

[BigBeachBanana]

Part b) plugin x=19 and solve for t.
[math]19 = 19 - 14e^{-0.3t}[/math]

 

[Dr.Peterson]

Part (b) asks us to find to the nearest degree, the time taken for the liquid to reach ambient temperature.

I cannot see how they arrive at an answer of approximately 11 minutes?

That makes no sense. Time is in minutes, not degrees. What is the exact wording?

More important, Newton's Law of Cooling implies it will never reach ambient temperature.

After 11 minutes, I get 18.28 degrees. But when I solve for when it will be 18.5 degrees (so that it will round to 19 to the nearest degree), I get 11.1 minutes, so maybe that's what they want, rounded to the nearest minute. Odd problem!

 

[BigBeachBanana]

I couldn't understand how they got the constant of proportionality of 0.3
[math]\frac{dx}{dt}=5.7=K(19-5) \implies K\approx .407[/math]Unless
[math]\frac{dx}{dt}=5.7=K(19-0) \implies K=.3[/math]But this assumes the temperate is at [imath]0\degree ?[/imath]

 

[Dr.Peterson]

I couldn't understand how they got the constant of proportionality of 0.3
[math]\frac{dx}{dt}=5.7=K(19-5) \implies K\approx .407[/math]Unless
[math]\frac{dx}{dt}=5.7=K(19-0) \implies K=.3[/math]But this assumes the temperate is at [imath]0\degree ?[/imath]

I hadn't looked yet at part (a) itself. You're right; k should be not 0.3, but 5.7/14 = 0.4. Everyone, including the OP, seems to be assuming the equation is right and solving it, rather than proving it, which is what was asked. They seem to have confused x with t.

I also notice the the "initial rate of cooling" as stated in the problem is wrong; it's warming up!

We really need to see the exact original question; we know from post #5 that it wasn't quoted properly (the nearest degree) wasn't mentioned).