Question

From Table \(14.4,\) we know that approximately \(2256 \mathrm{kJ}\) are needed to transform \(1.00 \mathrm{kg}\) of water at \(100^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\). What is the change in entropy of \(1.00 \mathrm{kg}\) of water evaporating at \(100.0^{\circ} \mathrm{C} ?\) (Specify whether the change in entropy is an increase, \(+\), or a decrease, \(-.\) )

Short answer

Answer: The change in entropy is an increase of approximately 6040.12 J/(K⋅kg).

Step-by-step solution

Identify the given values

We are given the following information: - Energy \(Q = 2256 \mathrm{kJ/kg} \times 1000 \mathrm{J / kJ} = 2256000 \mathrm{J / kg}\) (Converted kJ to J for convenience) - Mass of water \(m = 1.00 \mathrm{kg}\) - Temperature \(T = 100^{\circ} \mathrm{C}\)

Convert temperature to Kelvin

We need to convert the temperature from Celsius to Kelvin before we can use it in our formula. The conversion is: \(T_\mathrm{k} = T_\mathrm{c} + 273.15\) where \(T_\mathrm{k}\) is the temperature in Kelvin and \(T_\mathrm{c}\) is the temperature in Celsius. Thus, \(T_\mathrm{k} = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\)

Calculate the change in entropy

Now, we can use the formula to calculate the change in entropy for this process: \(\Delta S = \frac{Q}{T}\) Substituting our values: \(\Delta S = \frac{2256000 \mathrm{J / kg}}{373.15 \mathrm{K}}\) \(\Delta S \approx 6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\) Since heat is absorbed during the evaporation process, the change in entropy will be positive: \(\Delta S = +6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\) Thus, the change in entropy of \(1.00 \mathrm{kg}\) of water evaporating at \(100.0^{\circ} \mathrm{C}\) is an increase of approximately \(6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\).