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Rate of decrease

  • Thread starter Gwilim
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  • #1
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This question was in the exam I just sat, it was of a type I hadn't practiced, and I'd like someone to check my answer, as if it's wrong then I'm almost certain to have failed.

At 7 a.m. I made a cup of tea; after adding some milk it is about 90 C. When I left at 7:30 a.m. the tea is still drinkable at about 40 C. When I get back home at 8 a.m. the tea has cooled to 30 C.

Assume that the rate of cooling of the tea is proportional to the temperature difference of the teas and the temperature of the house. Assume also that the temperature of the house is constant.

What is the temperature of the house?

My answer: 27.5 C
 

Answers and Replies

  • #2
Tom Mattson
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How about showing your work? It would make it a lot easier for us to spot a wrong turn if you would show us the route you took!
 
  • #3
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Hmm I only got to take the question paper home, not the answer paper. I started with putting dT/dt = k(T-h) where h was the temperature of the house. Then I seperated the variables, integrated and took the exponentials which gave me (T-h) = ce^kt where c is a constant. Then I plugged in the values given in the question which gave me a set of silmultaneous equations to solve, and I ended up having to form a quadratic in e^k which gave me k = 0 and k = ln 1/5. I took k = 1/5 with the reason supplied that k < 0. This also supplied me with a value for c, namely 62 1/2. Plugging those values into my set of linear equations gave me h = -27.5 and from there I gave an argument centred around the fact that I should have been using |T-h| in place of (T-h) as to why h = 27.5
 
  • #4
tiny-tim
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Plugging those values into my set of linear equations gave me h = -27.5 and from there I gave an argument centred around the fact that I should have been using |T-h| in place of (T-h) as to why h = 27.5
Hi Gwilim! :smile:

Yes, 27.5 is right! :smile:

You'll lose a few marks for the |T- h| stuff … it should have worked out fine with (T - h).

(And you could have got a linear equation in (90 - h) instead of a quadratic in k, if you'd just squared one of the e^kt equations and subtracted it from the other).

But you've definitely got most of the marks for that question!

I hope the others were as good! :smile:
 

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