suppose a sample of a substance is heated in a container of boiling water until its temperature is 100° C. The sample is cooled by being left in a room of 24° C. The value a equals the difference between the sample’s initial temperature and the surrounding temperature. The value c represents the surrounding temperature. The cooling constant of the substance is k = 0.12. Find the sample’s temperature after 20 minutes.
Newton’s Law of Cooling is y = ae^(-kt) + c. The formula relates a substance’s temperature, y, to the time, t, in minutes that it has spent cooling.

Question

suppose a sample of a substance is heated in a container of boiling water until its temperature is 100° C. The sample is cooled by being left in a room of 24° C. The value a equals the difference between the sample’s initial temperature and the surrounding temperature. The value c represents the surrounding temperature. The cooling constant of the substance is k = 0.12. Find the sample’s temperature after 20 minutes.
Newton’s Law of Cooling is y = ae^(-kt) + c. The formula relates a substance’s temperature, y, to the time, t, in minutes that it has spent cooling.

anonymous · Answer

this is not as complicated as it looks, since you are told all the numbers you need to compute this

anonymous · Answer

well not exactly, you need $a$ 
$k=.12, t=20$ and $a=100-24=76$

anonymous · Answer

compute $$\large76\times e^{-.12\times 20}+24$$with a calculator

anonymous · Answer

assuming the "cooling constant" is given in terms of minutes