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A science Question. Stuck

V

VTboy

Banned
You take a termometer that has reach equilibrium with a room at constant temperature. Into a new room that has a constant temperature of 5 degrees C. After 1 min the thermometer reads 55 degrees and after 5 mins (or 4 min after the initial 1 min reading) it reads 30 degrees. What was the initial temp. I get around 64 degrees is that right.
 
If the temp decrease is linear, then the equation of the line describing the data is

y = -6.25x + 61.25, where 'y' is temp and 'x' is time. You want the temp at time = 0 (or in other words, x = 0), and solving for that you get y = 61.25, i.e. the temp of the first room was a constant 61.25 °C.
 
The answer depends on what level of science class you're taking.
AFAIK, it's not a linear function, but could probably be treated as one in lower level science classes.
I'm looking for the actual function now...
 
i think u use newton's law of cooling. where dT/dt is proportional to difference in teperature

another question from your diff eq. class?
 
Newton's law of cooling. I think it follows the form T = Ce^kt, but i'm drawing a blank on where to go from there.
 
Agreed... Newton's law of cooling. Rate of cooling is proportional to the difference in temperature between the object and the surrounding medium.


edit: Use the equation above, you've got 2 sets of data. Solve for C and k. Plug in time = 0 to find the original temperature. Final solution: next time read the thermometer before you leave the room.
 
Yes it was from my DE Class. I used dT/dt=k(T-Tm)

Solved the DE to get T-Tm=ce^(kt)

I then pluged stuff in to get

50=ce^k
25=ce^(5k)

I solved for c and k

Then Pluged them in and solved for T. To get around 64.46 Does that sound right.
 
Yes it was from my DE Class. I used dT/dt=k(T-Tm)

Solved the DE to get T-Tm=ce^(kt)

I then pluged stuff in to get

50=ce^k
25=ce^(5k)

Then I solved for c. Then I plug stuff into T-Tm=ce^(kt) with k being 0 for the initial temperature and with that I solved for T.

Then Pluged them in and solved for T. To get around 64.46 Does that sound right.
 
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