# Entropy: gas heated by resistor

1. Sep 3, 2016

### JMatt7

1. The problem statement, all variables and given/known data
A current $I=0.2 A$ flows in a resistor $R = 50 Ω$ immersed in a rigid adiabatic vessel that contains $n=3$ moles of Helium. The initial temperature of the system is $T_0 = 27 °C$. The resistor has a mass $m = 10 g$ and specific heat $c = 0.2 (cal/K)/g$ and runs for $t= 10 min$. Find:
1. the internal energy change of He.
2. the entropy change of He and the entropy change of the Universe.

2. Relevant equations
$P = i^2R$
$dS = \left( \frac {\delta Q} T \right)$

...

3. The attempt at a solution
I think I solved the problem, but I'm not completely sure of my reasoning, especially on the last question. Could you tell me what you think?

1) Helium is a noble gas, so I assumed it's an ideal gas. Therefore $U = U(T)$.
Since the volume is kept constant: $ΔU = n c_V ΔT = n \frac 3 2 RΔT$. (Helium is monoatomic)
I calculated the temperature at equilibrium ($T_{eq}$) by knowing that the electrical energy is converted into heat that heats up both the resistor and the gas:
$$E_{in} = (mc + n c_v) (T_{eq} - T_0)$$
$$I^2 R t + T_0 (mc + n c_v) = T_{eq} (mc + n c_v)$$
and hoping I converted all units accordingly:
$$T_{eq} = \frac {I^2 R t + T_0 (mc + n c_v)} {(mc + n c_v)} ≈ 431 K$$
Therefore: $$ΔU = n \frac 3 2 RΔT ≈ 4.89 kJ$$

2) Helium absorbs heat at constant volume, so $TdS =\delta Q = nc_v dT$
So: $$ΔS_{He} = \int_{T_0}^{T_{eq}} \frac {nc_v} T \, dT = n c_v \ln{\frac {T_{eq}} {T_0}} ≈13.5 J/K$$
As for the entropy change of the universe, I calculated the entropy change due to the heating of the resistor and added it to that of the gas:
$$ΔS_R = \int_{T_0}^{T_{eq}} \frac {mc} T \, dT = mc \ln{\frac {T_{eq}} {T_0}} ≈3.02 J/K$$
$$ΔS_{univ} = ΔS_{He} + ΔS_R ≈ 16.52 J/K$$

Since the energy flowing into the system comes only through electrical work, and the vessel is adiabatic, there are no other heat exchanges that could cause a change in entropy of the universe, right?

2. Sep 3, 2016

### TSny

Hello and welcome to PF!

Your method and resoning look very good to me. However, I'm getting different numbers.

For example, you have
The formula looks correct, but I don't get 431 K. Can you tell us what numerical values you have for $mc$ for the resistor and $nc_v$ for the He?

3. Sep 3, 2016

### JMatt7

Sure. I might have been a bit sloppy with numbers. So using the given data:
$mc = 10 g ~ 0.2~ \frac {cal} {K~g} = 2~ \frac {cal} K$
$nc_v = 3~ mol~ \frac {3}{2}~ R = 3~mol~ \frac {3}{2}~ 1.987~ \frac {cal}{K ~mol} ≈ 8.94 \frac {cal} K$
Therefore, since $1 ~cal = 4.18 ~J$:
$$T_{eq} = \frac {6000 ~J + 300.15~ K (2+8.94)~ 4.18 \frac J K} {(2+8.94)~ 4.18 \frac J K} ≈ \frac {6000 ~J + 13727.5 ~J } {45.74 \frac J K} ≈ 431.3 K$$

4. Sep 3, 2016

### TSny

OK. The problem appears to be with your calculation of $I^2Rt$. Did you square the current?

5. Sep 4, 2016

### JMatt7

Ahaha. Yes, you're right. Apparently I didn't. So $I^2Rt = 1200 ~J$ and therefore:
$T_{eq} ≈ 326.4~ K$
$ΔU ≈ 982.1 ~J$
$ΔS_{He} ≈ 3.14 ~J/K$
$ΔS_R ≈ 0.7 ~J/K$

6. Sep 4, 2016

### TSny

OK. I get about 987 J for ΔUHe, but that's close to what you get.