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Srathmore
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I am currently on exercise 10C in the edescel book, question number 4.

Newton's Law of cooling leads to the result θ=Ae-kt where θ is the excess of temperature of the cooling body over its surroundings at time t. For a given liquid, k=1/4, θ is measured in oC and t is measured in minutes. THe surroudning temperature is 20 oC and at t=, the temperature of the liquid is 60oC.

a)Find A
b)Find an expression for the temperature, ToC, of the liquid at time t
c)Calculate the temperature of the liquid when t=4
d)Find the time taken for the liquid to cool to 30oC
e)Find the rate of cooling of the liquid when t=2.

Any help would be appreciated. Please can you explain the question too. It seemed really long winded, and i don't really know what they want.

I did (a) and got it right. (60-20)=Ae-0k
Therefore A=40.

Thanks again for your help
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SunGod87
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Okay so now you know that
θ = 10e^(-1/4t)
Calculate the temperature when t = 4
Substitute t = 4
θ = 10e^(-1)

---------------

30 = 10e^(-1/4t)
3 = e^(-1/4t)
ln3 = -1/4t
(-4)(ln3) = t

Gotta eat dinner now I'll try the last part if noone has by the time I get back.

Basically it's a case of putting numbers in, nothing too long winded.
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Library
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(Original post by SunGod87)
Okay so now you know that
θ = 10e^(-1/4t)
Calculate the temperature when t = 4
Substitute t = 4
θ = 10e^(-1)

---------------

30 = 10e^(-1/4t)
3 = e^(-1/4t)
ln3 = -1/4t
(-4)(ln3) = t

Gotta eat dinner now I'll try the last part if noone has by the time I get back.

Basically it's a case of putting numbers in, nothing too long winded.
That's wrong. You can't just sub t as 4 because you havent got part (b) out. And where did you get A=10 from?
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SunGod87
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Wow, I didn't read the question properly at all, I assumed theta was temperature.

And A = 10 comes from the original poster posting it was 10.
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Library
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#5
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Nah, this is a diff Q.

How would you do b?
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