Need help with simple cooling problem.

  • Thread starter xsgx
  • Start date
  • #1
29
0
I just need someone to confirm my answer so I can be certain that my book has a typo and I am not making a some minuscule mistake that is leading me to the wrong answer.

A piece of metal is heated to 300 degrees Celsius and then placed in a cooling liquid at 50 degrees Celsius. After 4 minutes, the metal has cooled to 175 degrees Celsius. Find it's temperature after 12 minutes.



Equation: F(t)= To+ Ce^-kt

e= Euler's number

I started out solving for k


175= 50 + 300e^-k4
-50 -50

125=300e^-k4
/300 /300
-The Celsius units cancel out-

125/300=e^-k4
Ln Ln

Ln(125/300)= -4k
/-4 /-4

k= .21887

then I plug in the value of k and solve for what temperature the metal will be after 12 minutes have passed.

F(t)= 50+300e^-.21887*12

= 71.700

I end up getting the answer above but the book says the answer is 81.25

Thanks in advance for your help.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,213
7,299
I just need someone to confirm my answer so I can be certain that my book has a typo and I am not making a some minuscule mistake that is leading me to the wrong answer.

A piece of metal is heated to 300 degrees Celsius and then placed in a cooling liquid at 50 degrees Celsius. After 4 minutes, the metal has cooled to 175 degrees Celsius. Find it's temperature after 12 minutes.



Equation: F(t)= To+ Ce^-kt

e= Euler's number

I started out solving for k


175= 50 + 300e^-k4
-50 -50

125=300e^-k4
/300 /300
-The Celsius units cancel out-

125/300=e^-k4
Ln Ln

Ln(125/300)= -4k
/-4 /-4

k= .21887

then I plug in the value of k and solve for what temperature the metal will be after 12 minutes have passed.

F(t)= 50+300e^-.21887*12

= 71.700

I end up getting the answer above but the book says the answer is 81.25

Thanks in advance for your help.
If you plug t=0 into your first equation, does it give the right initial temperature?
 
  • #3
29
0
If you plug t=0 into your first equation, does it give the right initial temperature?
Wow.. I can't believe I missed that. Thank you.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,213
7,299
Wow.. I can't believe I missed that. Thank you.
Btw, there's a very quick way to the answer in this case. You're given an initial temperature difference of 250C, and a later difference of 125C, so you know how long it takes for the difference to halve. In a further twice that time, it quarters to 31C.
 

Related Threads on Need help with simple cooling problem.

Replies
1
Views
1K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
1K
Replies
14
Views
6K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
Top