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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

For a more realistic estimate of the maximum coefficient of performance of a heat pump, assume that a heat pump takes in heat from outdoors at $10^{\circ} \mathrm{C}$ below the ambient outdoor temperature, to account for the temperature difference across its heat exchanger. Similarly, assume that the output must be \(10^{\circ} \mathrm{C}\) hotter than the house (which itself might be kept at \(20^{\circ} \mathrm{C}\) ) to make the heat flow into the house. Make a graph of the coefficient of performance of a reversible heat pump under these conditions as a function of outdoor temperature (from $\left.-15^{\circ} \mathrm{C} \text { to }+15^{\circ} \mathrm{C} \text { in } 5^{\circ} \mathrm{C} \text { increments }\right)$
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
(a) What is the entropy change of 1.00 mol of \(\mathrm{H}_{2} \mathrm{O}\) when it changes from ice to water at \(0.0^{\circ} \mathrm{C} ?\) (b) If the ice is in contact with an environment at a temperature of \(10.0^{\circ} \mathrm{C}\) what is the entropy change of the universe when the ice melts?
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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