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Chapter 20: Problem 2

What is the magnitude of the change in entropy when \(6.00 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed to water at \(100{ }^{\circ} \mathrm{C} ?\) a) \(46.6 \mathrm{~J} / \mathrm{K}\) c) \(36.3 \mathrm{~J} / \mathrm{K}\) b) \(52.4 \mathrm{~J} / \mathrm{K}\) d) \(34.2 \mathrm{~J} / \mathrm{K}\)

Short Answer

Expert verified
Answer: c) \(36.3 \mathrm{~J} / \mathrm{K}\)

Step by step solution

01

Identify the known values

We know the mass of steam (m) is 6.00 g, the temperature (T) in °C is 100°C, and the latent heat of vaporization for water (L) is 2260 J/g. First, we need to convert the temperature to Kelvin by adding 273.15 to 100°C: 100°C + 273.15 = 373.15 K.
02

Calculate the change in entropy

Using the formula for entropy change during a phase transition: ∆S = mL/T, we can calculate the change in entropy as follows: ∆S = (6.00 g)(2260 J/g) / (373.15 K)
03

Simplify and solve for ∆S

Simplify the expression and solve for ∆S: ∆S = 6.00 * 2260 / 373.15 ≈ 36.3 J/K
04

Compare the result to the given options

Our calculated change in entropy is approximately 36.3 J/K, which corresponds to option c. Therefore, the correct answer is: c) \(36.3 \mathrm{~J} / \mathrm{K}\)

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Most popular questions from this chapter

Why might a heat pump have an advantage over a space heater that converts electrical energy directly into thermal energy?

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