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Chapter 20: Problem 52

To heat 1 cup of water \(\left(250 \mathrm{~cm}^{3}\right)\) to make coffee, you place an electric heating element in the cup. As the water lemperature increases from \(20^{\circ} \mathrm{C}\) to \(78^{\circ} \mathrm{C}\), the temperature of the heating element remains at a constant \(120^{\circ} \mathrm{C}\). Calculate the change in entropy of (a) the water, (b) the heating element; (c) the system of water and heating element. (Make the same assumption about the specific heat of water as in Example 20.10 in Section \(20.7,\) and ignore the heat that flows into the ceramic coffee cup itself.) (d) Is this process reversible or irreversible? Explain.

Short Answer

Expert verified
The entropy change of water will be calculated in Step 1, and of the heating element in Step 2. The total entropy of the system is the sum of these two values as detailed in Step 3. The determination of whether the process is reversible or irreversible is tackled in Step 4. Exact values depend on calculation. Note: The thermal capacity or mass of the heating element, and thus its energy transfer capability, is not provided; therefore detailed calculations or an exact answer can't be given.

Step by step solution

01

Entropy change of water

First, let's find the change in entropy of water. To do this, apply the formula for entropy change. The specific heat capacity of water at constant pressure \(C_p\) is approximately 4.186 Joule/gram/Kelvin. Convert the volume of water to mass (since 1cm^3 of water weighs approximately 1 gram), resulting in \(m = 250\) grams. Apply these values to the entropy change formula: \(\Delta S_{water} = \int_{293}^{351}\frac{250*4.186d_T}{T}\) (293 K and 351 K correspond to 20掳C and 78掳C respectively). Integrate this function to find the entropy change of water.
02

Entropy change of heating element

The heating element maintains a constant temperature, specifically 120掳C or 393 K. Thus the entropy change of the heating element is zero since there is no change in its temperature.
03

Entropy change of the system

The total entropy change of the system is the sum of the entropy changes of its parts. Since the entropy change of the heating element is zero, the entropy change of the system is essentially the entropy change of the water. Hence, \(\Delta S_{system} = \Delta S_{water}\). Now we have the entropy change for the entire system.
04

Reversibility check

A process is reversible if the total entropy change is equal to zero. If it's not zero, then the process is irreversible. Check if \(\Delta S_{system}\) is equal to 0 or not to answer this part of the problem.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a principle that describes the conservation of energy in any process involving heat and work. According to this law, energy cannot be created or destroyed; only transferred or converted from one form to another. In simpler terms:
  • Energy added to a system increases its internal energy or does work.
  • Energy removed decreases internal energy or extracts work.
In the context of heating water for coffee, electric energy is converted to thermal energy, which increases the water's internal energy. This warming of water is a practical example of The First Law of Thermodynamics ensuring that the energy supplied by the electric element results in a rise in water temperature.
Specific Heat Capacity
Specific heat capacity is a material's ability to absorb heat. It reflects how much heat energy a substance can take before its temperature changes by a specific amount. The specific heat capacity at constant pressure, noted as \(C_p\), for water is approximately 4.186 Joules per gram per Kelvin (J/g鈰匥).

In the exercise, we utilize the specific heat capacity of water to calculate the change in entropy. We use the fact that 1 cm鲁 of water weighs approximately 1 gram, resulting in a mass of 250 grams when we have 250 cm鲁 of water.
  • Energy required to increase the water's temperature from \(20\, ^{\circ}C\) (293 K) to \(78\, ^{\circ}C\) (351 K) is determined by this property.
  • The specific heat relation \(Q = mc \Delta T\) helps determine the energy involved.
  • Here, \(m = 250 \text{ g}\), \(c = 4.186 \text{ J/g鈰匥}\), and \(\Delta T = 58\, K\).
Specific heat capacity ensures that water is efficient at storing thermal energy due to its relatively high value compared to many other substances.
Irreversibility
Irreversibility refers to the characteristic of certain processes being unable to return to their original state without external influence. It is often related to increases in entropy, where the entropy change is greater than zero. An irreversible process involves net positive entropy generation, indicating energy spreads or disperses irreversibly.

In assessing reversibility, we determine that:
  • If entropy change (\(\Delta S\)) is not zero for the entire system, the process is irreversible.
  • When water is heated by the heating element, energy transfer cannot be entirely undone without additional work.
  • Due to the net entropy increase, we identify this process as irreversible.
In our case, the water heats by pulling energy from the heating element without simultaneous reverse change, posing a classical example of dissipative processes.
Entropy Change Calculation
Entropy measures the degree of disorder or randomness in a system and is key in determining the direction of thermodynamic processes. Entropy change, \(\Delta S\), can be calculated differently depending on whether the temperature changes or remains constant.

For the water in our scenario:
  • We use the formula \(\Delta S = \int \frac{dQ}{T}\) to find the entropy change.
  • Given constant pressure, use \(\Delta S_{water} = \int_{T_1}^{T_2}\frac{mC_p}{T}dT\) where \(m\) is the mass, \(C_p\) is specific heat, and \(T\) is absolute temperature.
  • Integrating from \(293K\) to \(351K\) using the specific heat capacity gives \(\Delta S_{water}\), which reflects the overall thermodynamic evolution during heating.
The heating element, however, maintains a constant temperature, so its entropy change is zero. The entire system's entropy, reflecting the sum of both entities鈥� entropy, illustrates both energy distribution and the system's irreversibility.

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Most popular questions from this chapter

Entropy of Metabolism. An average sleeping person metabolizes at a rate of about \(80 \mathrm{~W}\) by digcsting food or burning fat. Typically, \(20 \%\) of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body. where it is radiated away, The normal intemal temperature of the body (where the metabolism takes place) is \(37^{\circ} \mathrm{C},\) and the skin is typically \(7 \mathrm{C}^{3}\) cooler. By how much does the person's entropy change per sccond due to this heat transfer?

A Carnot engine whose high-temperature reservoir is at \(620 \mathrm{~K}\) takes in \(550 \mathrm{~J}\) of heat at this temperature in each cycle and gives up \(335 \mathrm{~J}\) to the low-temperature rescrvoir. (a) How much mechanical work does the cngine perform during cach cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal cfficicncy of the cycle?

The Outo-cycle engine in a Mercedes-Benz SL.K230 has a compression ratio of 8.8 . (a) What is the ideal efficiency of the engine? Use \(\gamma=1.40 .\) (b) The engine in a Dodge Viper GT2 has a slightly higher compression ratio of 9.6 . How much increase in the ideal efficicncy results from this increase in the compression ratio?

The theorctical cfficicncy of an Otto cycle is \(63 \% .\) The ratio of the \(C_{p}\) and \(C_{V}\) heat capacities for the gas in the cngine is \(\frac{7}{5}\). (a) What is the compression ratio \(r ?\) (b) If the net work done per cycle is \(12.6 \mathrm{~kJ}\), how much heat is rejected and how much is absorbed by the engine during each cycle?

You are conducting experiments to study prototype heat engines. In one test, \(4.00 \mathrm{~mol}\) of argon gas are taken around the cycle shown in Fig. \(\mathrm{P} 20.55 .\) The pressure is low cnough for the gas to be treated as ideal. You measure the gas temperature in states \(a, b, c,\) and \(d\) and find \(T_{a}=250.0 \mathrm{~K}\) \(T_{h}=300.0 \mathrm{~K}, \quad T_{c}=380.0 \mathrm{~K}, \quad\) and \(T_{d}=316.7 \mathrm{~K}\) (a) Calculate the efficiency \(e\) of the cycle. (b) Disappointed by the cycle's low efficicncy, you consider doubling the number of moles of gas while kecping the pressure and volume the same. What would \(e\) be then? (c) You remember that the efficiency of a Carnot cycle increases if the temperature of the hot reservoir is increased. So, you return to using \(4.00 \mathrm{~mol}\) of gas but double the volume in states \(c\) and \(d\) while keeping the pressures the same. The resulting temperatures in these states are \(T_{c}=760.0 \mathrm{~K}\) and \(T_{d}=633.4 \mathrm{~K}\). \(T_{a}\) and \(T_{b}\) remain the same as in part (a). Calculate \(e\) for this cycle with the new \(T_{c}\) and \(T_{d}\) values. (d) Encouraged by the increase in efficiency, you raise \(T_{c}\) and \(T_{d}\) still further. But \(e\) doesn't increase very much; it scems to be approaching a limiting valuc. If \(T_{a}=250.0 \mathrm{~K}\) and \(T_{b}=300.0 \mathrm{~K}\) and you keep volumes \(V_{a}\) and \(V_{b}\) the same as in part (a), then \(T_{d} T_{d}=T_{b} / T_{a}\) and \(T_{c}=1.20 T_{d}\). Derive an expression for \(e\) as a function of \(T_{d}\) for this cycle. What value docs \(e\) approach as \(T_{d}\) becomes very large?
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