Help with Thermodynamics Problem: Proving Energy & Calculating Heat Input

[Perses]

Hello, I was searching the web to find some information to help me solve a problem and i came across this forum. I was wondering if anyone would be able to help me with a homework problem I'm having.

On a cold day, where the outside temperature is 10degreesC, a room in a house is being heated from 10degrees to 20degrees. The volume of the room is 20m^3 and the atmospheric pressure is 101kPa. Cv = 3.6R
due to leaks in the room to the outside the pressure remains constant. I need to prove that the energy after the room has been heated is the same as the energy in the room before the heating.
What i did to solve this was calculate the energy before and the energy after and showed that they were equal.. can someone recommend a better path to prove this?

The next part of the question asks me to calculate the heat input to reach 20degrees in the room at this constant pressure.
For this; i thought i could use dQ = NCvdt + NRdt; however that answer is off from the answer provided in the back of the book by 5kJ or something to that extent(the answer is supposed to be 323kJ).. can anyone toss me a few hints?

Thanks a lot in advance!

 

[LeonhardEuler]

Keep in mind that as the room is heated, gas leaves. This means that N changes. Since we are apparently assuming the gas is ideal, you can say
N=\frac{PV}{RT}
Now the question becomes an integral.

 

[Andrew Mason]

Hello, I was searching the web to find some information to help me solve a problem and i came across this forum. I was wondering if anyone would be able to help me with a homework problem I'm having.

On a cold day, where the outside temperature is 10degreesC, a room in a house is being heated from 10degrees to 20degrees. The volume of the room is 20m^3 and the atmospheric pressure is 101kPa. Cv = 3.6R
due to leaks in the room to the outside the pressure remains constant. I need to prove that the energy after the room has been heated is the same as the energy in the room before the heating.

PV=nRT

PV is the internal energy of the gas. If P and V remain constant then so does PV and nRT. n_1RT_1 = n_2RT_2. This means that as the room temperature increases, n decreases.

AM

 

[Perses]

Ah, thank you guys very much. I had thought that the moles had changed, but i didn't think about taking the integral; i was instead trying to find dQ.. how stupid of me. Thanks again!

 

[Andrew Mason]

Ah, thank you guys very much. I had thought that the moles had changed, but i didn't think about taking the integral; i was instead trying to find dQ.. how stupid of me. Thanks again!

You do have to find dQ = dU + dW. This consists of the amount of heat needed to raise the air temperature of the air in the room by dT plus the amount of work done in pushing a volume of air dV out.

For an incremental change in T (dT) there is an incremental change in volume dV = nRdT/P. Since n = PV/RT, dV = VdT/T.

So dQ = dU+dW = nC_vdT + PdV

Substituting n = PV/RT:

dQ = PVC_vdT/RT + PVdT/T = PV(1 + \frac{C_v}{R})dT/T [/itex]<br /> <br /> So to find the total amount of heat added one has to integrate dQ: <br /> \int_{T_1}^{T_2} dQ = (1+\frac{C_v}{R})PV\int_{T_1}^{T_2} dT/T = (1+\frac{C_v}{R})PVln(\frac{T_2}{T_1})<br /> <br /> AM