A thermometer reading 4˚C is brought into a room where the temperature reading is 30˚C. If the thermometer reads 10˚C after 2 minutes, determine the temperature reading 5 minutes after the thermometer is first brought into the room.
A thermometer reading 4˚C is brought into a room where the temperature reading is 30˚C. If the thermometer reads 10˚C after 2 minutes, determine the temperature reading 5 minutes after the thermometer is first brought into the room.
I have no idea what this all means. Where have you used Newton's law of cooling/warming to set up the differential equation? Where is your working for solving that differential equation?
Please show ALL the work leading to your answer if you want your answer reviewed.
So you didn't actually use "Newton' law of cooling", you used a formula derived from that.
Okay, here $\displaystyle H_0= 4$, $\displaystyle H_s= 30$ and your formula becomes $\displaystyle H= (4- 30)e^{kt}$. You want to find H when t= 5 but you still don't know what "k" is. You can find that by using the fact that H= 10 when t= 2: $\displaystyle 10= (4- 30)e^{2k}$. Then you can find H when t= 5 by using that same k and t= 5.
Warning: This is NOT linear! H(5) is NOT equal to H(2)+ H(3)! Do NOT find H(3), just use the formula with t= 5, not 3.
k is a constant. It does not change.
Were you expected to use Newton's Law of Cooling to solve this question? If yes, then the fact that k is a constant should be clear and you shouldn't be taking shortcuts by applying formulas you don't understand. If not, then the title of your post is misleading and a lot of time might have been wasted answering this question in a way that would ultimately have made no sense to you.