(a) An object is placed in a 68°F room. Write a differential equation for H, the temperature of the object at time t.
ANSWER: dH/dt = -k(68 - H)

(b) Give the general solution for the differential equation.
ANSWER: H = 68 - Ce-kt

(c) The temperature of the object is 40°F initially, and 48°F one hour later. Find the temperature of the object after 3 hours.
ANSWER: 57.8°F

I got the answers for (a) and (b) correct, but I have no idea what I'm doing wrong for C. The answer is from the book. If anyone can look at my work below and tell me how to correct it in clear terms (I'm not at all good with math; I'm terrible at it, in fact), then I'd be really grateful.

68 - H = Be-kt
68 - 40 = Be-k(0) = B = 28 (trying to find the constant)
68 - 48 = 28e-k(1)
20 = 28e-k(1)
ln(20/28) = ln(e-k)
-0.3365 approx. = -k so 0.3365 approx. = k
H = 68 - 28e0.3365(3)

And we can tell it's been wrong since I've been trying to find B, but then I have no idea how to find the correct value or if I'm going about it the completely wrong way. Thanks ahead of time for any help!


@Ross already answered but I want to add that there is another way.

We know that the difference between the body's temperature and the surrounding decreases exponentially. So if in the beginning it was 28 and after an hour it was 20, then we can conclude that the difference every hour gets smaller by a factor of $\frac{20}{28}=\frac{5}{7}$.

So after 3 hour the difference will be smaller by $\left (\frac{5}{7} \right ) ^3 $, meaning the temperature will be:

$$ 68 - 28\cdot \left (\frac{5}{7} \right ) ^3 = \frac{2832}{49} = 57.7959... $$


You dropped the minus sign in the exponent in the last line. You were doing fine up to there. I get $57.7959^\circ$ F after $3$ hours (counting from when it was $40^\circ$ F)

  • $\begingroup$ In H = 68 - 28e^0.3365(3)? I thought the two negatives cancelled, so I dropped it... Unless that's not what you do? Thanks for clarifying, by the way! $\endgroup$ – bri kayyy Sep 6 '13 at 3:22
  • $\begingroup$ Yes, $k$ is positive as you show in the line before. But there is a negative sign in the equation in the first line, so it should be $H=68-28e^{-0.3365*3}$ $\endgroup$ – Ross Millikan Sep 6 '13 at 3:35

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