Question

when a hot object is placed in a water bath whose temp is 25C, it cools from 100C to 50C in 170s. in another bath, the same cooling occurs in 150s. Find the temp of the second bath

Answer 1

How can you start this?

Answer 2

don't know how to

Answer 3

Maths of Heat transfer?

Answer 4

no idea

Answer 5

looks like newtons law of cooling, right?

Answer 6

yes

Answer 7

what a pain this is, but i think we can do it

Answer 8

The function for cooling models exponential decay:
\[T(t) = C e^{-kt} + T_f\]
Tf is final temp or temp of water bath ---> Tf = 25
Initial temp = C + Tf ----> C = 100 -25 = 75
\[T(t) = 75e^{-kt} + 25\]
Use given example to find constant "k"
T(170) = 50
\[50 = 75e^{-170k} +25\]
\[k = \frac{\ln 3}{170}\]
Now using this k value you can find temp of 2nd water bath
T(150) = 50
\[50 = (100-T_f)e^{-\ln 3 *(150/170)} + T_f\]
\[T_f = \frac{50 -100(3^{-15/17})}{1 - 3^{-15/17}} \approx 19.44\]
This means 2nd bath is cooler at 19.44C since it takes less time to cool object to 50C

Answer 9

@dumbcow
Is not C equal to 75 as you mentioned in the first case (170 seconds) C = 100 -25 = 75?

I think it is a constant in the equation and does not change due to the case!

Answer 10

@3mar
The C value changes in 2nd case because it is dependent on temp of water bath.
It was really a bad use of the constant C, I should have written it as
\[T(t) = (T_0 - T_f) e^{-kt} + T_f\]
which makes the 2nd case equation more clear.
Sorry for the confusion

Answer 11

No problem.
That makes sense now!

It more clearer now.
Thank you for good explanation, @dumbcow.