# Irreversibility of a process

• Gregg
Q = mc(T2 - T1) ΔS = mc(T2 - T1) So, ΔQ is the same as ΔS. Both are negative (due to the absorption of heat by the sea).

## Homework Statement

A mass m with heat capacity c at temperature $T_1$ is dropped into the sea at $T_2$.

What is the irreversibility of the process?

## Homework Equations

$$I = T_0 (\Delta S + \Delta S_E )$$

$$I = W_{\text{rev}} - W$$

## The Attempt at a Solution

Heat in is $Q=mc\Delta T$ the temperature of the sea remains constant by the first law W=Q. What is the distinction between the work for a reversible process and the work?

No work is involved. ΔW =0, ΔU = -Q for the mass and +Q for the sea.

U of the sea does change, albeit mightily little.

Write an expression for the entropy change of the mass and of the sea.

What constitutes the irreversibility calls for a verbal answer, plus reference to the line above.

Well since the mass of the sea is not given I assume that the change in internal energy is negligable. Made an error with work! Of course there is no work.

So the key is to find the entropy change of the sea and of the block.

$$\Delta s_{\text{sea}} = \frac{mc(T_2-T_1)}{T_2}$$

$$\Delta s_{\text{block}} = ?$$

is it log?

No, U of the sea goes up. It's an infinite mass all right, so an infinitely small rise in temperature, but ΔU and Q go up together & are finite. Where do you think the loss of Q from the mass goes? Of course, ΔT of the sea is essentially zero. Have to be careful with infinities and zeros.

Gregg said:
$$\Delta s_{\text{block}} = ?$$

is it log?
Yes. You have dS = dQ/T and dQ=mc dT. Put those together and integrate.

$I = T_0(mc(\frac{T_0-T_1}{T_0})+mc \log(\frac{T_0}{T_1}))$

$T_1 > T_0.$ So, is $\Delta S_{\text{sea}} < 0$ ?

I know that $mc \log (\frac{T_0}{T_1}) < 0$

No, the sea absorbs heat, so its entropy increases.

Anyway, you got ΔQ and ΔS for the mass correct now. Although, simplify by writing ΔQ of mass = mc(T2 - T1).

So what do you think ΔQ of the sea is? And then ΔS of the sea?