Heat Problem

[Jason76]

A flask in running water is kept at the constant temperature of 50 degrees. After 1, minute the temperature inside the flask drops to 80 degrees. After 2 minutes from the beginning of the cooling process, the temperature inside the flask is 70 degrees. What was the original temperature inside the flask?


\(\displaystyle H - H_{S} = (H_{0} - H_{S})e^{-kt}\) where \(\displaystyle H_{0}\) is the temperature at \(\displaystyle t = 0\)

Solving for \(\displaystyle H_{0}\)

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

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

\(\displaystyle 70 = (50) + (H_{0} - (50))e^{-k(2)}\) - Find k, but two variables here.

 

[Ishuda]

A flask in running water is kept at the constant temperature of 50 degrees. After 1, minute the temperature inside the flask drops to 80 degrees. After 2 minutes from the beginning of the cooling process, the temperature inside the flask is 70 degrees. What was the original temperature inside the flask?


\(\displaystyle H - H_{S} = (H_{0} - H_{S})e^{-kt}\) where \(\displaystyle H_{0}\) is the temperature at \(\displaystyle t = 0\)

Solving for \(\displaystyle H_{0}\)

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

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

\(\displaystyle 70 = (50) + (H_{0} - (50))e^{-k(2)}\) - Find k, but two variables here.

Writing it a slightly different way
\(\displaystyle H_0 = H_{S} +(H - H_{S})e^{kt}\)
So
\(\displaystyle H_0 = H_{S} +(80 - H_{S})e^{k}\)
\(\displaystyle H_0 = H_{S} +(70 - H_{S})e^{2k}\)
or subtracting
\(\displaystyle 0 = (80 - 50)e^{k} - (80 - 50)e^{2k}\)
Now you could let x = e^k and solve the quadratic equation but I suggest there is an easier way. [EDIT: Oh, and fix that dumb mistake]

 

[Jason76]

Writing it a slightly different way
\(\displaystyle H_0 = H_{S} +(H - H_{S})e^{kt}\)
So
\(\displaystyle H_0 = H_{S} +(80 - H_{S})e^{k}\)
\(\displaystyle H_0 = H_{S} +(70 - H_{S})e^{2k}\)
or subtracting
\(\displaystyle 0 = (80 - 50)e^{k} - (80 - 50)e^{2k}\)
Now you could let x = e^k and solve the quadratic equation but I suggest there is an easier way. [EDIT: Oh, and fix that dumb mistake]

What happened to t?

 

[Ishuda]

What happened to t?

At t=1 H is 80, at t=2 H is 70, use those to find K as indicated. Then substitute back to find H₀

 

[Jason76]

Solving for \(\displaystyle H_{0}\)

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

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

 

[Ishuda]

Solving for \(\displaystyle H_{0}\)

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

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

What is k? Follow the steps outlined above.