Heat Problem - # 2

[Jason76]

A hard boiled egg at \(\displaystyle 97^{o} C\) is put in a sink of \(\displaystyle 17^{o} C\)water. After 5 min, the egg's temperature is \(\displaystyle 37^{o} C\) Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach \(\displaystyle 19^{o} C\)?

\(\displaystyle H - H_{S} = (H_{0} - H_{S})e^{-kt}\)

\(\displaystyle H = H_{S} + (H_{0} - H_{S})e^{-kt}\)

\(\displaystyle H = (17) + ((97) - (17))e^{-kt}\)

\(\displaystyle H = (17) + (80)e^{-kt}\)

find k

\(\displaystyle 37 = (17) + (80)e^{-k(5)}\)


\(\displaystyle 20 = (80)e^{-k(5)}\)

\(\displaystyle \dfrac{1}{4} = e^{-k(5)}\)

\(\displaystyle \ln(\dfrac{1}{4}) = \ln(e^{-k(5)})\)

\(\displaystyle -1.386294361 = -k(5)\)

\(\displaystyle k = 0.277258872\)

\(\displaystyle 19 = (17) + (80)e^{-(0.277258872)(t)}\)


\(\displaystyle 2 = (80)e^{-(0.277258872)(t)}\)

\(\displaystyle \dfrac{1}{4} = e^{-(0.277258872)(t)}\)

\(\displaystyle \ln[\dfrac{1}{4}] = \ln[e^{-(0.277258872)(t)}]\)

\(\displaystyle -1.386294361 = -(0.277258872)(t)\)

\(\displaystyle t = 0.199999999\) This doesn't look right.

 

[Ishuda]

Start at this point
\(\displaystyle 2 = (80)e^{-(0.277258872)(t)}\)
then
\(\displaystyle \frac{1}{40} = e^{-(0.277258872)(t)}\)
NOT
\(\displaystyle \frac{1}{4} = e^{-(0.277258872)(t)}\)

 

[HallsofIvy]

A hard boiled egg at \(\displaystyle 97^{o} C\) is put in a sink of \(\displaystyle 17^{o} C\)water. After 5 min, the egg's temperature is \(\displaystyle 37^{o} C\) Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach \(\displaystyle 19^{o} C\)?

\(\displaystyle H - H_{S} = (H_{0} - H_{S})e^{-kt}\)

\(\displaystyle H = H_{S} + (H_{0} - H_{S})e^{-kt}\)

\(\displaystyle H = (17) + ((97) - (17))e^{-kt}\)

\(\displaystyle H = (17) + (80)e^{-kt}\)

find k

\(\displaystyle 37 = (17) + (80)e^{-k(5)}\)


\(\displaystyle 20 = (80)e^{-k(5)}\)

\(\displaystyle \dfrac{1}{4} = e^{-k(5)}\)

Notice that, for this problem, it is sufficient to know that \(\displaystyle e^{-k}= \dfrac{1}{4^{5}}\). The value of k itself is not needed.

\(\displaystyle \ln(\dfrac{1}{4}) = \ln(e^{-k(5)})\)

\(\displaystyle -1.386294361 = -k(5)\)

\(\displaystyle k = 0.277258872\)

\(\displaystyle 19 = (17) + (80)e^{-(0.277258872)(t)}\)

Knowing that \(\displaystyle e^{-k}= \dfrac{1}{4^{1/5}}\), this is
\(\displaystyle 19= 17+ \dfrac{80}{4^{t/5}}\)

\(\displaystyle 2 = (80)e^{-(0.277258872)(t)}\)

\(\displaystyle \dfrac{1}{4} = e^{-(0.277258872)(t)}\)

\(\displaystyle \ln[\dfrac{1}{4}] = \ln[e^{-(0.277258872)(t)}]\)

\(\displaystyle -1.386294361 = -(0.277258872)(t)\)

\(\displaystyle t = 0.199999999\) This doesn't look right.