Thermodynamics B/C

kickit · Post by **kickit** » May 6th, 2018, 11:04 am

Hello,
Does anyone know the answer to this question? I have no idea how to solve it, and I don't know calculus.

Two identical rooms are at 15 C and 25 C respectively. They are separated by an aluminum wall of thickness 0.5 m and area 10 m2. Assume the rooms are isolated from the rest of the world. The thermal conductivity coefficient of aluminum is 237 W m-1K-1

(4 Points) After a long time, what temperature will the rooms come to, at equilibrium? (For this part, assume the aluminum wall in between the rooms is negligible)

(20 Points) How long will it take for the rooms to come within 5 C of each other? Show work.

MattChina · Post by **MattChina** » May 6th, 2018, 11:09 am

kickit wrote:Hello,
Does anyone know the answer to this question? I have no idea how to solve it, and I don't know calculus.

Two identical rooms are at 15 C and 25 C respectively. They are separated by an aluminum wall of thickness 0.5 m and area 10 m2. Assume the rooms are isolated from the rest of the world. The thermal conductivity coefficient of aluminum is 237 W m-1K-1

(4 Points) After a long time, what temperature will the rooms come to, at equilibrium? (For this part, assume the aluminum wall in between the rooms is negligible)

(20 Points) How long will it take for the rooms to come within 5 C of each other? Show work.

well for the first one it would just 20 degrees, and for the second one u would use the equation of heat and the temperature difference would be 5.

knightmoves · Post by **knightmoves** » May 6th, 2018, 12:03 pm

kickit wrote:Hello,
Does anyone know the answer to this question? I have no idea how to solve it, and I don't know calculus.

Two identical rooms are at 15 C and 25 C respectively. They are separated by an aluminum wall of thickness 0.5 m and area 10 m2. Assume the rooms are isolated from the rest of the world. The thermal conductivity coefficient of aluminum is 237 W m-1K-1

(4 Points) After a long time, what temperature will the rooms come to, at equilibrium? (For this part, assume the aluminum wall in between the rooms is negligible)

(20 Points) How long will it take for the rooms to come within 5 C of each other? Show work.

How can you solve this second part without knowing the heat capacity of the rooms?

kickit · Post by **kickit** » May 6th, 2018, 12:25 pm

Sorry, the heat capacity is 100 kJ/K.

Post by **UTF-8 U+6211 U+662F** » May 6th, 2018, 12:51 pm

MattChina wrote:
kickit wrote:Hello,
Does anyone know the answer to this question? I have no idea how to solve it, and I don't know calculus.

Two identical rooms are at 15 C and 25 C respectively. They are separated by an aluminum wall of thickness 0.5 m and area 10 m2. Assume the rooms are isolated from the rest of the world. The thermal conductivity coefficient of aluminum is 237 W m-1K-1

(4 Points) After a long time, what temperature will the rooms come to, at equilibrium? (For this part, assume the aluminum wall in between the rooms is negligible)

(20 Points) How long will it take for the rooms to come within 5 C of each other? Show work.
for the second one u would use the equation of heat and the temperature difference would be 5.

Can you elaborate on this

Alex-RCHS · Post by **Alex-RCHS** » May 6th, 2018, 1:12 pm

My solution:

$dQ/dt = -kAT/x$
where $dQ/dt$ is the rate of heat flow, $k$ is the thermal conductivity, $A$ is the area, $T$ is the difference in temperature between the two rooms, and $x$ is the thickness of the separation.

You also know that, for 2 degrees that the difference in temperature of the two rooms increases, 100,000 J of heat is transferred (when the temperature difference increases by 2 degrees, it means one room gained 1 degree and the other lost one degree, and based on the specific heat capacity you get the 100,000 J). So:
$dQ=50,000dT$

Which means that:
$-kAT/x=50,000dT/dt$
Plugging in numbers, $-kA/x=-4740$ , so:

$-.0948T=dT/dt$

T is probably $Ae^(^B^t^)$ , where $B=-.0948$ . By plugging in $t=0$ , $T=10$ , we get $A=10$ .

So I got this function: $T=10e^(^-^.^0^9^4^8^t^)$ .

Plug in T=5, and you get t=7.312 seconds...

kickit · Post by **kickit** » May 6th, 2018, 1:15 pm

Thank you, it seems like the right answer.

Alex-RCHS · Post by **Alex-RCHS** » May 6th, 2018, 1:22 pm

kickit wrote:Thank you, it seems like the right answer.

Did you get this from a practice test? What was on the answer key?

kickit · Post by **kickit** » May 6th, 2018, 1:24 pm

Yeah, I got it from a practice test, and the answer key says that you are right.

Justin72835 · Post by **Justin72835** » May 6th, 2018, 2:19 pm

kickit wrote:Hello,
Does anyone know the answer to this question? I have no idea how to solve it, and I don't know calculus.

Two identical rooms are at 15 C and 25 C respectively. They are separated by an aluminum wall of thickness 0.5 m and area 10 m2. Assume the rooms are isolated from the rest of the world. The thermal conductivity coefficient of aluminum is 237 W m-1K-1

(4 Points) After a long time, what temperature will the rooms come to, at equilibrium? (For this part, assume the aluminum wall in between the rooms is negligible)

(20 Points) How long will it take for the rooms to come within 5 C of each other? Show work.

Here's an alternate solution! One thing I love about calc problems is how versatile they are in how you can approach them.

My Work (Click to Show Hidden Text)

[img]https://i.imgur.com/NOxAV7B.jpg[/img]
Also, sorry for using n-solve at the very end lol.

Scioly.org

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