How Do You Calculate Initial Temperature Using Newton's Law of Cooling?

AI Thread Summary
To calculate the initial temperature using Newton's Law of Cooling, the equation T(room) + (T(initial) - T(room)) e^(-kt) is applied. In this scenario, the room temperature and final temperature are both 10 degrees Celsius, leading to confusion about the initial temperature. The calculations suggest that if the object cooled to room temperature in 180 seconds, it must have started at room temperature, which contradicts the premise. Alternative initial temperatures, such as 100 degrees Celsius, yield final temperatures around 24.9 degrees Celsius, indicating that the initial data may be incorrect. The discussion emphasizes the importance of accurate initial conditions for solving cooling problems effectively.
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A hot object cools down to a temperature of 10 degrees of Celsius during a period of 180 seconds. The room temperature is 10 degrees of Celsius. Assume the coefficient in the Newton's Law of cooling to be 0.01[1/sec]. Determine the original temperature of the substance



2. T(room) + (T(initial) T(room)) e^-kt = T(final)

t room = 10
ti = ?
Tfinal = 10

3. 10 + (Ti -10 ) e^(-0.01)(180) = 10
10 + Ti - 10 + 0.1652 = 10
0.1652 = 10 = 60.53

I think this wrong because its a multiple choice and the choices are 93 and 100

Please help me thank you!
 
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Spinnor said:
Shouldn't T final be a little above room temperature? Anyway a problem similar to this is done here,

http://ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

See example 1.

Also you could plug in your possible answers and see if they work, I don't think they will as I think your initial data is off?

:( Sorry I still struggling with the problem, can you please show me how to do ? please!
 
You wrote,

"10 + (Ti -10 ) e^(-0.01)(180) = 10"

Subtract 10 from both sides and we have,

(Ti -10 ) e^(-0.01)(180) = 0 --> Ti = 10The object did not cool down to room temperature in 180 seconds unless it started at room temperature. I think your facts are wrong. Assume that 100 was the starting temp and work backward to find the final temp, which can't be 10.Tf = 10 + (100 - 10)*exp(-1.8) Tf = 24.9

Tf = 10 + (93 - 10)*exp(-1.8) Tf = 23.7

Good luck!

You might also visit,

http://demonstrations.wolfram.com/NewtonsLawOfCooling/
 
Last edited:
sorry, what is the ambient temperature? I've assignment where the question as follow:

the temperature T of a cooling object drops at a rate proportional to the difference T-S, where S is constant temperature of surrounding medium. If initially T= 100 C, find the temperature of the cooling object at any time. anyone...please help me..
 
fahanaam, refer to the rules of this forum.
 
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