HELP! HND Maths Assignment ("The rate of cooling for the new room...")

[wilksefc]

Hello All,

i was wondering if somebody might be able to help with the below question...I don't want the answer, more the help to work it out for myself. I was off college when this was covered so i'm quite stuck!
Thanks for your help in advance...

YOU NEED TO CONFIRM THE FOLLOWING:
The rate of cooling for the new room when the heat source has been turned off,
The trending Newtons law of cooling is given via the following equation:

Temp Difference (θ) = 15e<sup>-0.035t

Where</sup> θ is temp diff t = time.

The client needs you to confirm the rate of cooling and how long it will take for the internal temp to cool below 15oC.
Internal temp setpoint is19oC.
External ambient is 2.68oC

 

[HallsofIvy]

I assume that the "temperature difference" here is the difference between the internal temperature at any time and the external temperature which is the constant 2.68 degrees Celcius. So, given the formula for "temperature difference" at time t, \(\displaystyle T= 15e^{-0.035t}\), the internal temperature is \(\displaystyle 15e^{-0.035t}+ 2.68\). When will that be equal to 15?

 

[wilksefc]

I assume that the "temperature difference" here is the difference between the internal temperature at any time and the external temperature which is the constant 2.68 degrees Celcius. So, given the formula for "temperature difference" at time t, \(\displaystyle T= 15e^{-0.035t}\), the internal temperature is \(\displaystyle 15e^{-0.035t}+ 2.68\). When will that be equal to 15?

thanks