/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Water is boiled at \(100^{\circ}... [FREE SOLUTION] | 91Ó°ÊÓ

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Water is boiled at \(100^{\circ} \mathrm{C}\) and \(1.0\) atm. Under these conditions, 1.0 g of water occupies \(1.0 \mathrm{~cm}^{3}, 1.0\) g of steam occupies 1670 \(\mathrm{cm}^{3}\), and \(L_{v}=540 \mathrm{cal} / \mathrm{g}\). Find \((a)\) the external work done when \(1.0\) g of steam is formed at \(100^{\circ} \mathrm{C}\) and \((b)\) the increase in internal energy.

Short Answer

Expert verified
(a) 169.05 J, (b) 2096.31 J

Step by step solution

01

Define the Work Done Formula

When a substance changes its phase, external pressure is performed by the substance through expansion. The work done by the system during this expansion can be defined by the formula: \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume of the system.
02

Calculate the Change in Volume

The change in volume \( \Delta V \) is the volume of the steam minus the volume of the liquid water. Using the given values, \( \Delta V = 1670 \ \mathrm{cm}^{3} - 1.0 \ \mathrm{cm}^{3} = 1669 \ \mathrm{cm}^{3} \).
03

Convert Units for Pressure

The pressure is given as \( 1.0 \) atm. Since 1 atm = 101.3 J/L and there are 1000 cm³ in a liter, we have \( 1.0 \ \text{atm} = 101.3 \, \frac{J}{1000\,\mathrm{cm}^{3}} \).
04

Calculate Work Done

The work done by the steam, \( W \), is calculated as: \[ W = P \Delta V = \left( 101.3 \, \frac{J}{1000\,\mathrm{cm}^{3}} \right) \cdot 1669 \,\mathrm{cm}^{3} \approx 169.0477 \ \text{Joules} \].
05

Calculate Heat Absorbed (Using Latent Heat)

The heat absorbed, \( Q \), when 1 g of water converts to steam can be calculated using the latent heat: \( Q = L_v \times m = 540 \, \text{cal/g} \times 1.0 \, \text{g} = 540 \, \text{cal} \). Convert 540 cal to Joules using 1 cal = 4.184 J. Thus, \( Q = 2265.36 \ \text{Joules} \).
06

Calculate Increase in Internal Energy

The increase in internal energy (\( \Delta U \)) is given as the difference \( Q - W \). From Step 5 and Step 4, \( \Delta U = 2265.36 \ \text{J} - 169.05 \ \text{J} \approx 2096.31 \ \text{J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Change
When a substance transforms from one state of matter to another, it's known as a phase change. In our context, water is changing from a liquid to a gaseous state. This is called vaporization or boiling. This transition occurs at a specific temperature and pressure, in this case, at \(100^{\circ} \mathrm{C}\) and \(1.0\) atm. During a phase change, energy is added or removed from the system without altering its temperature, ensuring the substance undergoes a transformation.
  • Melting and Freezing: Phase change between solid and liquid.
  • Vaporization and Condensation: Phase change between liquid and gas.
Understanding phase changes helps us address how energy inputs impact the state of a material. For water, this is crucial since it allows us to calculate the work done and changes in energy when it turns into steam.
Latent Heat
Latent heat is the energy absorbed or released by a substance during a phase change at a constant temperature and pressure. Unlike sensible heat, which changes the temperature of a substance, latent heat does not alter the temperature but changes its state instead.
  • Latent Heat of Fusion: The heat required to convert a solid into a liquid without temperature change.
  • Latent Heat of Vaporization: The heat required to convert a liquid into a gas without temperature change.
In the given problem, the latent heat of vaporization \(L_v\) is \(540 \mathrm{cal} / \mathrm{g}\). This implies that each gram of water absorbs 540 calories to become steam at \(100^{\circ} \mathrm{C}\), facilitating the phase change without a change in temperature.
Internal Energy
Internal energy is the total energy contained within a system due to its particles' motion and position. It includes kinetic and potential energies at a microscopic scale. In the context of phase changes, internal energy is influenced by the heat absorbed and the work done by the system.
To find the change in internal energy during phase changes, we use the formula:\[\Delta U = Q - W\]where \(\Delta U\) is the change in internal energy, \(Q\) is the heat absorbed, and \(W\) is the work done by the system. For the system in question, converting 1 g of water to steam results in an internal energy increase of approximately 2096.31 Joules. This reflects the energy required for the phase transition plus the work done against the atmospheric pressure.
Work Done
Work done is a concept from physics that describes the energy transferred by a force over a distance. In thermodynamics, it represents the energy required for a system to expand or contract.
In the scenario of boiling water turning into steam, work is done by the system when the volume of water vapor increases. To calculate work done, we use the formula:\[ W = P \Delta V \]where \(P\) is the external pressure (1 atm) and \(\Delta V\) is the change in volume (1670 cm³ - 1 cm³). The work needed for this phase change, calculated in the solution, is approximately 169.05 Joules. This \(W\) signifies the energy transferred to push against the atmospheric pressure, allowing the water to expand into steam.

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Most popular questions from this chapter

Compute the maximum possible efficiency of a heat engine operating between the temperature limits of \(100^{\circ} \mathrm{C}\) and \(400{ }^{\circ} \mathrm{C}\). Remember that our thermodynamic equations are expressed in terms of absolute temperature. The most efficient engine is the Carnot engine, for which $$ \text { Efficiency }=1-\frac{T_{\mathrm{L}}}{T_{\mathrm{H}}}=1-\frac{373 \mathrm{~K}}{673 \mathrm{~K}}=0.446=44.6 \% $$

By how much does the internal energy of 50 g of oil \((c=0.32 \mathrm{cal} / \mathrm{g}\) \({ }^{\circ} \mathrm{C}\) ) change as the oil is cooled from \(100^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).

Twenty cubic centimeters of monatomic gas at \(12{ }^{\circ} \mathrm{C}\) and 100 \(\mathrm{kPa}\) is suddenly (and adiabatically) compressed to \(0.50 \mathrm{~cm}^{3}\). Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, \(P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}\) where \(\mathrm{Y}=1.67\) for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\). Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$ $$ \begin{aligned} \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000 &=7000 \end{aligned} $$

If a person does \(8.00 \mathrm{~h}\) of moderate physical labor "burning" 400 kcal/h, by how much does his or her internal energy change as a result?

A 100-g bullet \(\left(c=0.030 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) is initially at \(20{ }^{\circ} \mathrm{C}\). It is fired straight upward with a speed of \(420 \mathrm{~m} / \mathrm{s}\), and on returning to the starting point strikes a cake of ice at exactly \(0^{\circ} \mathrm{C}\). How much ice is melted? Neglect air friction.

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