# Thread: writing equation for integration regarding rate of change

1. ## writing equation for integration regarding rate of change

Hi

An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

Any ideas would be appreciated

2. Originally Posted by Poppy
Hi

An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

Any ideas would be appreciated
I suggest using the model $\frac{dT}{dt} = k (90 - T)$ where $T(5) = 70$.

3. ## formular

I also posted this on another website and the reply was the same equation except it was dT/dt=k(T-90) not dT/dt=k(90-T). Could you please explain why yours is correct, I am a little confused haha

Thanks

4. Originally Posted by Poppy
I also posted this on another website and the reply was the same equation except it was dT/dt=k(T-90) not dT/dt=k(90-T). Could you please explain why yours is correct, I am a little confused haha

Thanks
k represents two different things in the two DE's. In mine k > 0 (why?), in the other one k < 0 (why?).

5. Originally Posted by mr fantastic
k represents two different things in the two DE's. In mine k > 0 (why?), in the other one k < 0 (why?).
Is it because the object gains energy as opposed to losing it (as in the original Newtons equation)? 

Anyway, my integration does not seem to be right either

dT/dt =k(90-T)
dT =k(90-T) dt
1/(90-T) dT=k dt

-log(90-T)=kt+C
log(90-T)=-kt0+C
90-T=e(-kt)+C
T=-e(-kt)+90+C

70-90=-e(-k(5))+C
log(20)/-5=k+C

29-90=-e(-k(0))+C
64=-1+C
65=C

(log(20)/-5)-65=k
k=-65.59914645

T=-e(65.59914645(10))+65
T=-3.0856546...*10^28

This cannot be right...where am i going wrong?

6. ## my e ing was wrong

-log(90-T)=kt+C
log(90-T)=-kt0+C
90-T=e(-kt)+C
T=-e(-kt)+90+C

becomes

-log(90-T)=kt+C
log(90-T)=-kt-C
90-T=e(-kt-C)
T=-e(-kt)*e(-C)+90 // e(-C)=K
T=-Ke(-kt)+90

7. Originally Posted by Poppy
Is it because the object gains energy as opposed to losing it (as in the original Newtons equation)?

Anyway, my integration does not seem to be right either

dT/dt =k(90-T)
dT =k(90-T) dt
1/(90-T) dT=k dt

-log(90-T)=kt+C
log(90-T)=-kt0+C
90-T=e(-kt)+C Mr F says: You don't seem to realise that C is also meant to be in the exponent .... (Errrmmm, you do realise that e is raised to an exponent, don't you ....? Because what you have written suggests multiplication). Everything that follows will be wrong. Also, and not that it has consequences here, modulus signs should have been used when integrating to get the log: it should have been log |...| not log(....)

T=-e(-kt)+90+C

70-90=-e(-k(5))+C
log(20)/-5=k+C

29-90=-e(-k(0))+C
64=-1+C
65=C

(log(20)/-5)-65=k
k=-65.59914645

T=-e(65.59914645(10))+65
T=-3.0856546...*10^28

This cannot be right...where am i going wrong?
I'd say a lot of your trouble is handling the basic algebra of exponentials. You might be wise to review that material.

sorry, my fault those were typos- well i typed them intentionally that way but thinking they ment other things(that is the way they are typed into matlab) haha
Thanks for all your other help
i managed to do the problem

9. Originally Posted by Poppy
Hi

An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

Any ideas would be appreciated
Heat always moves from the hotter body to the cooler. Whether heat moves from the object to the water or "the other way around" depends entirely upon whether the object or the water is hotter. You can use "Newton's law of cooling" in either case.