- #1

tezktenr

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## Homework Statement

The temperature of a 12-oz (0.354-L) can of soft drink is reduced from 20 to 5 ºC by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at 20 ºC to the surroundings. There are no other heat transfers. Determine the minimum theoretical work input required by the cycle, in kJ, assuming the soft drink is an incompressible liquid with the properties of liquid water. Ignore the aluminum can.

## Homework Equations

ΔU = Qnet - Wnet

ΔS = Qnet/T + σ

## The Attempt at a Solution

After learning general concepts of 2nd law of thermo, I recently just started to learn entropy in thermodynamics. I am so confused with this new concept and don't know what it actually means so that I am not able to apply it in the questions.

To solve this problem, basically what I think was to regard it as a reversible cycle to get the ideal minimum work input for this refrigeration cycle.

For a refrigeration cycle, β = Q

_{L}/ (Q

_{H}- Q

_{L}) = Q

_{L}/ W

_{in}.

Also, for an ideal reversible cycle, β = T

_{L}/ (T

_{H}- T

_{L}).

I think the hot and cold reservoirs for this refrigeration cycle is

T

_{L}= 5 + 273.15 = 278.15 K

T

_{H}= 20 + 273.15 = 293.15 K

After substituting all those known temperature, I get β = 18.543.

Now, my idea is to get the value of Q

_{L}. Then, I can find the minimum work input of a refrigeration cycle.

To do that, I only look at the soft drink (incompressible liquid) system.

ΔU = Qnet - Wnet.

Looking at a system that only consists of the liquid, heat (Q

_{L}) was taken from it. However, the liquid is incompressible so that the volume does not change. If the volume of the liquid does not change, Wnet for this system should be zero since Wnet = ∫P*dV.

Thus,

ΔU = - Q

_{L}

So, I can calculate ΔU to find Q

_{L}.

ΔU = m * c

_{p}* (T

_{2}- T

_{1})

T

_{2}= T

_{H}= 293.15 K;

T

_{1}= T

_{L}= 278.15 K;

m = V/v = 0.354 * 10

^{-3}m

^{3}/ 10

^{-3}m

^{3}/kg

c

_{p}= 4.2.

So, what am I doing wrong?

I knew I have to use entropy but I don't see where to use it. I have the solution.

It uses the equation: ΔS = Qnet/T + σ. To compute the minimum work input, it assumes σ = 0. (That is, assuming it is a reversible process). So far, I understood.

Then, Qnet = ΔS * T + σ * T (In which T = 293.15 and I don't understand why using this temperature here either)

I understand how to calculate ΔS but the following is what I am mainly confused about.

The solution calculates the work by using: Work minimum = Qnet - ΔU.

Since ΔU is the change of internal energy of the liquid (soft drink),

why would the work calculated here be the input work of the refrigeration cycle?

I think the input work of the refrigeration cycle is not directly applied to the liquid but instead to some other components that could help remove heat from the liquid. Am I Wrong? What concepts am I misunderstanding?

Shouldn't this work here be zero because the volume of the liquid doesn't change?