Question

Question: (I) What is the change in entropy of 320 g of steam at 100°C when it is condensed to water at 100°C?

Short answer

The change in entropy of the steam is \( - 1939\;{\rm{J/K}}\).

Step-by-step solution

Understanding entropy

Entropy is the function of the state of a system which is measure of the order or disorder of a system.

When heat Q is added to a system by a reversible process, at a constant temperature T, then change in entropy of the system is

\(\Delta S = \frac{Q}{T}\).

However, if the heat is lost from the system, then Q is taken negative.

Given information

The mass of steam is \(m = 320\;{\rm{g}} = 320 \times {10^{ - 3}}\;{\rm{kg}}\).

Temperature is\(T = 100^\circ {\rm{C}} = \left( {100 + 273} \right)K = 373\;{\rm{K}}\).

Latent heat of vaporization is \({L_{\rm{v}}} = 22.6 \times {10^5}\;{\rm{J/kg}}\).

Determination of entropy of the system

When steam gets condenses to water at \(100^\circ {\rm{C}}\), it releases heat. The heat released per unit mass of steam at this constant temperature is termed the latent heat of vaporization.

So, the total amount of heat released is

\(\begin{aligned}{l}\frac{Q}{m} &= {L_{\rm{v}}}\\Q &= m{L_{\rm{v}}}.\end{aligned}\)

Since heat is released from the system, the change in entropy of the steam is

\(\begin{aligned}{c}\Delta S &= - \frac{Q}{T}\\ &= - \frac{{m{L_{\rm{v}}}}}{T}.\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}\Delta S &= - \frac{{\left( {320 \times {{10}^{ - 3}}\;{\rm{kg}}} \right) \times 22.6 \times {{10}^5}\;{\rm{J/kg}}}}{{373\;{\rm{K}}}}\\ &= - 1939\;{\rm{J/K}}\end{aligned}\)

Thus, the change in the entropy of the steam is \( - 1939\;{\rm{J/K}}\).