When 2.000 kg of water evaporates, 22.60 × 10⁵ J/kg of heat are needed. What is the entropy change of the water in the boiling process?

Question

Quizplus · Accepted Answer

The entropy change of the water in the boiling process can be calculated using the formula:

ΔS = Q/T

where ΔS is the entropy change, Q is the amount of heat added or removed, and T is the temperature.

In this case, Q is given as:

Q = m × h

where m is the mass of water evaporated and h is the heat of vaporization.

So, Q = 2,000 kg × 22.60 × 10^6 J/kg = 45.20 × 10^9 J

We also need the temperature at which the water boils. This is given by the boiling point of water at atmospheric pressure, which is 100°C or 373 K.

Substituting these values into the formula, we get:

ΔS = Q/T = (45.20 × 10^9 J)/(373 K) = 121,202 J/K

Therefore, the entropy change of the water in the boiling process is 121,202 J/K, which is closest to choice A, 12,120 J/K.

When 2000 Kg of Water Evaporates, 22