# Newton's law of cooling

1. Nov 11, 2007

### kring_c14

at 1 pm, a thermometer reading 70 F is taken outside where the air temperature is -10F (ten below zero). at 1:02 p.m., the reading is 26F. At 1:05 p., the thermometer is taken back indoors , where the air is at 70 F. What is the temperature reading at 1:09 pm?

Ive made a table like this
Tm= -10

T---- 70 --- 26 --- x

t---- 0 --- 2--- 5

am I supposed to get the temperature x, the do this??
Tm=70

T ---- x ---- n

t ---- 0 ---- 4

Last edited: Nov 11, 2007
2. Nov 11, 2007

### kring_c14

I was absent the day this was discussed..I would really appreciate some help..thank you//

3. Nov 11, 2007

### kring_c14

the answer in the book is 56 F but I couldn't get it right..

4. Nov 11, 2007

### kring_c14

heeeelllppp..

5. Nov 11, 2007

### hage567

Do you know the equation you need for this? Check your book. What you need to do first is use it to find k, the constant that describes the rate of cooling.

6. Nov 11, 2007

### kring_c14

yes, it is ln(T-Tm)]$$^{T2}_{T1}$$=kt]$$^{t2}_{t1}$$

Ive already computed the value of k..it is -0.40

this is what I did
ln(26-(-10))-ln(70-(-10))=k (2)
ln(36/80)=2k
k=-0.40

did I do it right?

7. Nov 11, 2007

### hage567

It shouldn't be negative, watch your signs.

8. Nov 11, 2007

### kring_c14

i dont know where Im wrong.. there was another problem on newton's law of cooling where I got a negative k.. but I got the right answer..

9. Nov 11, 2007

### hage567

This is the form I used

$$T(t) = T_a + (T_o +T_a) e^{-kt}$$ where $$T_a$$ is the ambient temperature, $$T_o$$ in the initial temperature, k is a constant, and t is time.

Well I got a positive one using the above equation, but keep going. You'll know if you get the correct answer.

Are you sure your equation is correct?

What do you think your next step should be?

10. Nov 12, 2007

### kring_c14

ok I got it already....lots of thanks!!!