Newton Law of Cooling

1. Dec 7, 2011

luigihs

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

2. Dec 7, 2011

3. Dec 8, 2011

luigihs

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

4. Dec 8, 2011

Spinnor

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 = 10

The 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: Dec 8, 2011
5. Dec 15, 2011

fahanaam

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..

6. Dec 16, 2011

LawrenceC

fahanaam, refer to the rules of this forum.