Entropy of running water around a resistor

AI Thread Summary
The discussion revolves around calculating the entropy changes in a resistor, running water, and the universe when a 10A current flows through a 20-ohm resistor at a constant temperature of 10°C. The participant notes that the resistor's entropy remains unchanged since the energy is work done rather than heat transfer. They calculate the power dissipated by the resistor as 2000W, equating to 2000J. The main challenge lies in correctly applying the entropy change formula while considering the constant temperature and converting temperature to Kelvin. Ultimately, they conclude that the entropy change for the universe is equivalent to that of the running water.
bmarson123
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Homework Statement


An electric current of 10A runs for 1 minute through a 20 ohm Resistor which is maintained at 10 degC by being immersed in running water. What are the entropy changes in a) the resistor b) the running water c) the universe


Homework Equations



P = I2R

dS = dQ/T

The Attempt at a Solution



For a) there is no change in the entropy as it is work done by the resistor, not heat.

b) I've calculated P = 2000W = 2000J

The problem I'm having is formulating the path between the initial and final states of the system because of the constant temperature. I'm also not sure if I'm using the right equation?

c) This will be the same as the answer for b) I think.
 
Physics news on Phys.org
Figured it out!

Forgot to convert the temperature to kelvin and then got everything written down in a mess!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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