Thermal Conductivity Problem-- Two Reservoirs

[Kara4566]

Suppose that a large mixture of ice and water is in thermal contact with a reservoir which contains a large quantity of water and steam. Assume that both the ice/water bath and the water/steam reservoirs are otherwise thermally isolated and at a constant temperature. The only contact between them is via a uniform rod which is made of a conductive metal and has a uniform cross-sectional area. The latent heat for the ice-water transition is 3.3×105 J kg , and the latent heat for the water-steam transition is 2.26×106 J kg .

Suppose that the rate of energy transport is 450W, that the thermal conductivity of the metal is 210 W m K , and that the rod is square with a side-length of 0.14m. What is the length of the rod?

What is the change in entropy of the water/steam mixture during a period where 1200J of energy is transported from the water/steam mixture to the ice/water mixture?The unknown variable I seem to require to solve both of the above questions is the temperature change between the reservoirs. Is there a way to find the temperature change using the latent heats given? I have attempted these problems numerous times but still, am not successful in finding the temperatures of the reservoirs. Are the temperatures even necessary to solve the problems? Any help is greatly appreciated!

 

[scottdave]

What temperature can water exist as both a solid (ice) and liquid at the same time?

Also think about the other phase state - gas (steam) alongside liquid water.

 

[trurle]

P=(Tsteam-Tice)*K*W*W/L
450=100*210*0.14*0.14/L
L=0.915m

entropy change is dH/Tice - dH/Tsteam = 1200/273 - 1200/373 = 4.396 - 3.217 = 1.179 J/K

 

[Kara4566]

P=(Tsteam-Tice)*K*W*W/L
450=100*210*0.14*0.14/L
L=0.915m

entropy change is dH/Tice - dH/Tsteam = 1200/273 - 1200/373 = 4.396 - 3.217 = 1.179 J/K

Thank you very much! Makes sense.

 

[scottdave]

As to whether you need to know the temperature difference, take a look at the dimensions of what you are given. Specifically, the thermal conductivity relates Power (Energy/Time) and Length and Temperature.

 

[Chestermiller]

P=(Tsteam-Tice)*K*W*W/L
450=100*210*0.14*0.14/L
L=0.915m

entropy change is dH/Tice - dH/Tsteam = 1200/273 - 1200/373 = 4.396 - 3.217 = 1.179 J/K

The question asks for the entropy change of the water-steam mixture, not the entropy change of the overall combined system.

 

[scottdave]

The question asks for the entropy change of the water-steam mixture, not the entropy change of the overall combined system.

But should the solution be even posted at all... right off the bat? The OP had not even a chance to attempt it. His original question was asking how to determine the temperatures of the two reservoirs.

 

[Chestermiller]

But should the solution be even posted at all... right off the bat? The OP had not even a chance to attempt it. His original question was asking how to determine the temperatures of the two reservoirs.

I think some gave the member a Warning for this.

 

[scottdave]

I think some gave the member a Warning for this.

That makes sense. I recall doing something similar when I was new.

 

[Chestermiller]

That makes sense. I recall doing something similar when I was new.

Who doesn't. Guilty too.