Newton's law of cooling and temperature

[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

 

[kring_c14]

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

 

[kring_c14]

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

 

[kring_c14]

heeeelllppp..

 

[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.

 

[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?

 

[hage567]

It shouldn't be negative, watch your signs.

 

[kring_c14]

i don't know where I am wrong.. there was another problem on Newton's law of cooling where I got a negative k.. but I got the right answer..

 

[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.

i don't know where I am wrong.. there was another problem on Newton's law of cooling where I got a negative k.. but I got the right answer..

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?

 

[kring_c14]

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