Question

One kilogram of water at 0°C is heated to 100°C. Compute its change in entropy. Assume that the specific heat of water is constant at 4190 J/kg·K over this temperature range.

Answer 1

The change in entropy for heating 1 kilogram of water from 0°C to 100°C is approximately 1300 J/K, calculated using the formula ΔS = m × cp × ln(T2/T1) with a specific heat capacity (cp) of 4190 J/kg·K.

Step-by-step explanation:

Calculating the Change in Entropy for Water from 0°C to 100°C

To calculate the change in entropy (ΔS) for 1 kilogram of water being heated from 0°C to 100°C, we use the formula:

ΔS = m × cp × ln(T2/T1)

Where:

• m is the mass of the water
• cp is the specific heat capacity of water
• T1 and T2 are the initial and final temperatures in Kelvin, respectively

Given:

• m = 1 kg (mass of the water)
• cp = 4190 J/kg·K (specific heat capacity of water)
• T1 = 273 K (0°C in Kelvin)
• T2 = 373 K (100°C in Kelvin)

Substitute the given values into the formula:

ΔS = 1 kg × 4190 J/kg·K × ln(373/273)

Calculate the natural logarithm:

ln(373/273) ≈ 0.3102

Therefore, the change in entropy is:

ΔS = 1 kg × 4190 J/kg·K × 0.3102 ≈ 1300.3 J/K

The change in entropy for 1 kilogram of water heated from 0°C to 100°C is approximately 1300 J/K.