- #1
JMatt7
- 7
- 1
Homework Statement
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:
- the internal energy change of He.
- the entropy change of He and the entropy change of the Universe.
Homework Equations
##P = i^2R##
##dS = \left( \frac {\delta Q} T \right)##
...
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?