When water is boiled at a pressure of 2.002.002.00 atm, the heat of vaporization is 2.20×1062.20\$$times10^62.20×106 J/kg $$and the boiling point is 120120120°C. At this pressure, 1.001.001.00 kg of water has a volume of 1.00×10−31.00\$$times10^{-3}1.00×10$$−3 m3, and 1.001.001.00 kg of steam has a volume of 0.8240.8240.824 m3. Compute the increase in internal energy of the water.

Ch 19: The First Law of Thermodynamics

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 19: The First Law of Thermodynamics

Problem 14b

Chapter 19, Problem 14b

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20$\times$10^6$ J/kg and the boiling point is $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00$\times$10^{-3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the increase in internal energy of the water.

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Identify the given values: pressure (P) = 2.00 atm, heat of vaporization (L) = 2.20 * 10^6 J/kg, initial volume of water (V1) = 1.00 * 10^-3 m^3, final volume of steam (V2) = 0.824 m^3, and mass (m) = 1.00 kg.

Convert the pressure from atm to pascals (Pa) using the conversion factor: 1 atm = 1.013 * 10^5 Pa. Therefore, P = 2.00 * 1.013 * 10^5 Pa.

Calculate the work done (W) during the expansion using the formula: W = P * (V2 - V1). Substitute the values for P, V2, and V1 to find the work done.

Use the first law of thermodynamics to find the change in internal energy (ΔU): ΔU = Q - W, where Q is the heat added to the system. Here, Q is equal to the heat of vaporization multiplied by the mass, Q = L * m.

Substitute the values of Q and W into the equation ΔU = Q - W to find the increase in internal energy of the water.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a temperature change. For water at 2.00 atm, this value is 2.20 * 10^6 J/kg. It is crucial for calculating the energy needed to vaporize water and understanding phase transitions.

First Law of Thermodynamics

The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed. It is expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. This principle helps calculate the change in internal energy when water boils.

Work Done by Expanding Gas

When a gas expands, it does work on its surroundings, calculated as W = PΔV, where P is the pressure and ΔV is the change in volume. In this scenario, the work done by the steam as it expands from liquid to gas at 2.00 atm is essential for determining the internal energy change during boiling.

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Textbook Question

When water is boiled at a pressure of $2.00$ atm, the heat of vaporization is $2.20 \times 10^{6}$ J/kg and the boiling point is $120$ °C. At this pressure, $1.00$ kg of water has a volume of $1.00 \times 10^{- 3}$ m³, and $1.00$ kg of steam has a volume of $0.824$ m³. Compute the work done when $1.00$ kg of steam is formed at this temperature.

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