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Chapter 15: Problem 49

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

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

Step by step solution

01

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 \(T = 100^{\circ} \mathrm{C}\)
02

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, \(T_\mathrm{k} = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\)
03

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 \(6040.12 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{kg}}\).

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Most popular questions from this chapter

A \(0.50-\mathrm{kg}\) block of iron $[c=0.44 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]\( at \)20.0^{\circ} \mathrm{C}$ is in contact with a \(0.50-\mathrm{kg}\) block of aluminum \([c=\) $0.900 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})]$ at a temperature of \(20.0^{\circ} \mathrm{C} .\) The system is completely isolated from the rest of the universe. Suppose heat flows from the iron into the aluminum until the temperature of the aluminum is \(22.0^{\circ} \mathrm{C}\) (a) From the first law, calculate the final temperature of the iron. (b) Estimate the entropy change of the system. (c) Explain how the result of part (b) shows that this process is impossible. [Hint: since the system is isolated, $\left.\Delta S_{\text {System }}=\Delta S_{\text {Universe }} .\right]$
A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.
An ideal gas is heated at a constant pressure of $2.0 \times 10^{5} \mathrm{Pa}\( from a temperature of \)-73^{\circ} \mathrm{C}$ to a temperature of \(+27^{\circ} \mathrm{C}\). The initial volume of the gas is $0.10 \mathrm{m}^{3} .\( The heat energy supplied to the gas in this process is \)25 \mathrm{kJ} .$ What is the increase in internal energy of the gas?
A student eats 2000 kcal per day. (a) Assuming that all of the food energy is released as heat, what is the rate of heat released (in watts)? (b) What is the rate of change of entropy of the surroundings if all of the heat is released into air at room temperature \(\left(20^{\circ} \mathrm{C}\right) ?\)
The efficiency of a muscle during weight lifting is equal to the work done in lifting the weight divided by the total energy output of the muscle (work done plus internal energy dissipated in the muscle). Determine the efficiency of a muscle that lifts a \(161-\mathrm{N}\) weight through a vertical displacement of \(0.577 \mathrm{m}\) and dissipates \(139 \mathrm{J}\) in the process.
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