10 gram of steam at 100 *C are blown of the surface of 90 g of water at 0*c,contained in calorimeter of water equivalent 10g,all the steam being condenced. Calcutate the increase in entropy of the system.

To calculate the increase in entropy of the system when steam is condensed into water and then mixed with cold water, we need to consider the heat exchanges that occur during this process. The system consists of steam at 100 °C, which will condense into water at the same temperature, and then this water will mix with the cold water at 0 °C. Let's break this down step by step.

Understanding the Components

We have the following components in our system:

• Steam: 10 grams at 100 °C
• Cold Water: 90 grams at 0 °C
• Calorimeter: Water equivalent of 10 grams

Calculating Heat Transfer

First, we need to calculate the heat released by the steam as it condenses and then cools down to 0 °C. The heat released during the condensation of steam can be calculated using the formula:

Q = m \times L

Where:

• Q: Heat released (in joules)
• m: Mass of steam (10 g)
• L: Latent heat of vaporization of water (approximately 2260 J/g)

Calculating the heat released during condensation:

Q_condensation = 10 g × 2260 J/g = 22600 J

Next, we need to calculate the heat lost by the condensed water as it cools from 100 °C to 0 °C:

Q_cooling = m \times c \times \Delta T

Where:

• c: Specific heat capacity of water (approximately 4.18 J/g°C)
• ΔT: Change in temperature (100 °C - 0 °C = 100 °C)

Calculating the heat lost during cooling:

Q_cooling = 10 g × 4.18 J/g°C × 100 °C = 4180 J

Total Heat Released

The total heat released by the steam as it condenses and cools down is:

Total Q = Q_condensation + Q_cooling = 22600 J + 4180 J = 26780 J

Calculating the Increase in Entropy

Now, we can calculate the increase in entropy for both the steam and the cold water. The entropy change (ΔS) can be calculated using the formula:

ΔS = Q / T

For the steam, the temperature during condensation is 100 °C (or 373 K):

ΔS_steam = Q_condensation / T = 22600 J / 373 K ≈ 60.6 J/K

For the cooling of the condensed water, we need to consider the average temperature during the cooling process. Since it cools from 100 °C to 0 °C, we can use the average temperature of 50 °C (or 323 K):

ΔS_cooling = Q_cooling / T_avg = 4180 J / 323 K ≈ 12.9 J/K

Combining Entropy Changes

The total increase in entropy for the system is the sum of the entropy changes:

Total ΔS = ΔS_steam + ΔS_cooling = 60.6 J/K + 12.9 J/K ≈ 73.5 J/K

Final Thoughts

The increase in entropy of the system, which includes the steam condensing and the subsequent cooling of the water, is approximately 73.5 J/K. This reflects the irreversible nature of the process and the overall increase in disorder as the steam transitions to water and mixes with the cold water.