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Chapter 4: Problem 2

A piston moves upward a distance of \(5 \mathrm{~cm}\) while \(200 \mathrm{~J}\) of heat is added (Fig. 4.23). Calculate the change in internal energy of the vapor if the spring is originally unstretched. SOLUTION The work needed to raise the weight and compress the spring is $$ \begin{aligned} W &=(m g)(h)+\frac{1}{2} K x^{2}+\left(P_{\mathrm{atm}}\right)(A)(h) \\ &=(60)(9.81)(0.05)+\left(\frac{1}{2}\right)(50000)(0.05)^{2}+(100000)\left[\frac{\pi(0.2)^{2}}{4}\right](0.05)=250 \mathrm{~J} \end{aligned} $$ The first law for a process without kinetic or potential energy changes is $$ Q-W=\Delta U $$ Thus, we have \(\Delta U=200-250=-50 \mathrm{~J}\).

Short Answer

Expert verified
The change in internal energy, \( \Delta U \), is -50 J.

Step by step solution

01

Determine the work done

Calculate the work needed to raise the weight, compress the spring, and account for atmospheric pressure: W = (m g)(h) + \frac{1}{2} K x^{2} + (P_{\text{atm}})(A)(h) Substituting the values: \[ m = 60 \text{ kg} \ g = 9.81 \text{ m/s}^2 \ h = 0.05 \text{ m} \ K = 50000 \text{ N/m} \ x = 0.05 \text{ m} \ P_{\text{atm}} = 100000 \text{ Pa} \ A = \frac{\text{蟺}(0.2^2)}{4} \ h = 0.05 \text{ m} \] Calculating each term: \[ (60)(9.81)(0.05) = 29.43 \text{ J} \ \ \left(\frac{1}{2}\right)(50000)(0.05^2) = 62.5 \text{ J} \ \ (100000)\left(\frac{蟺(0.2^2)}{4}\right)(0.05) = 158.05 \text{ J} \] Summing these terms: \[ W = 29.43 + 62.5 + 158.05 = 250 \text{ J} \]
02

Apply the first law of thermodynamics

Use the equation from the first law of thermodynamics for a system without kinetic or potential energy changes: \[ Q - W = \Delta U \] Where Q is the heat added to the system and W is the work done by the system. Given Q = 200 J, and W = 250 J: \[ \Delta U = 200 - 250 = -50 \text{ J} \]

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Key Concepts

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

Internal Energy
Internal energy is the total energy stored in a system. It includes both the microscopic kinetic energy, which comes from the movement of particles, and the potential energy due to interactions between particles. In thermodynamics, changes in internal energy are crucial to understanding how systems behave. When heat is added to a system or work is done on it, the internal energy changes. This concept is central to solving problems related to the First Law of Thermodynamics, as it determines the overall energy balance within the system.
Work Done
Work done in thermodynamics refers to energy transfer when a force moves an object. It's calculated based on the distance moved and the force applied. For example, when a piston moves, work is done to raise weights, compress springs, or displace the atmosphere.
  • **Mechanical Work**: Given by the formula \((mg)(h)\) where \(m\) is mass, \(g\) is gravitational acceleration, and \(h\) is height.
  • **Spring Work**: Calculated using \(\frac{1}{2}Kx^{2}\) where \(K\) is the spring constant and \(x\) is compression length.
  • **Atmospheric Work**: Involves pressure and area, given by \(P(A)(h)\) where \(P_{\text{atm}}\) is atmospheric pressure, \(A\) is area, and \(h\) is height.

Summing all these effects gives the total work done. In the given problem, the total work done is calculated as 250 J.
Heat Transfer
Heat transfer is the process of energy moving from one body or system to another due to temperature difference. In thermodynamics, heat (denoted as \(Q\)) is an essential form of energy transfer. It can enter or leave a system, affecting the internal energy.
  • **Added Heat**: If a system absorbs heat, energy moves from the surroundings into the system, increasing its internal energy.
  • **Released Heat**: Conversely, if heat is released, the system loses energy to the surroundings.

In the provided example, 200 J of heat is added to the system, initiating changes in internal energy and work done. Understanding this helps grasp how energy interacts and transforms within thermodynamic processes.
Energy Conservation
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed; it can only be transformed. This law is formulated as: \( Q - W = \Delta U \), where **Q** is the heat added, **W** is the work done, and **\Delta U** is the change in internal energy.

  • **Energy Balance**: When heat is added (\(Q\)), and work (\(W\)) is done by the system, the internal energy (\

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Most popular questions from this chapter

A \(10-\mathrm{m}^{3}\) tank is being filled with steam at \(800 \mathrm{kPa}\) and \(400^{\circ} \mathrm{C}\). It enters the tank through a \(10-\mathrm{cm}-d i a m e t e r\) pipe. Determine the rate at which the density in the tank is varying when the velocity of the steam in the pipe is \(20 \mathrm{~m} / \mathrm{s}\). SOLUTION The continuity equation with one inlet and no outlets is [see (4.56)]: $$ \rho_{1} A_{1} V_{1}=\frac{d m_{\text {c... }}}{d t} $$ Since \(m_{\text {c.v. }}=\rho V\), where \(V\) is the volume of the tank, this becomes $$ V \frac{d \rho}{d t}=\frac{1}{v_{1}} A_{1} V_{1} \quad 10 \frac{d \rho}{d t}=\left(\frac{1}{0.3843}\right)(\pi)(0.05)^{2}(20) \quad \frac{d \rho}{d t}=0.04087 \mathrm{~kg} / \mathrm{m}^{3} \cdot \mathrm{s} $$

A system undergoes a cycle consisting of the three processes listed in the table. Compute the missing values. All quantities are in \(\mathrm{kJ}\). \begin{tabular}{|c|c|c|c|} \hline Process & \(Q\) & \(W\) & \(\Delta E\) \\ \hline \(1 \rightarrow 2\) & \(a\) & 100 & 100 \\ \(2 \rightarrow 3\) & \(b\) & \(-50\) & \(c\) \\ \(3 \rightarrow 1\) & 100 & \(d\) & \(-200\) \\ \hline \end{tabular} Use the first law in the form \(Q-W=\Delta E\). Applied to process \(1 \rightarrow 2\), we have $$ a-100=100 \quad \therefore a=200 \mathrm{~kJ} $$ Applied to process \(3 \rightarrow 1\), the results are $$ 100-d=-200 \quad \therefore d-300 \mathrm{~kJ} $$ The net work is then \(\Sigma W=W_{1-2}+W_{2-3}+W_{3-1}=100-50+300=350 \mathrm{~kJ}\). The first law for a cycle demands that $$ \Sigma Q=\Sigma W \quad 200+b+100=350 \quad \therefore b=50 \mathrm{~kJ} $$ Finally, applying the first law to process \(2 \rightarrow 3\) provides $$ 50-(-50)=c \quad \therefore c=100 \mathrm{~kJ} $$ Note that, for a cycle, \(\Sigma \Delta E=0\); this, in fact, could have been used to determine the value of \(c\) : $$ \Sigma \Delta E=100+c-200=0 \quad \therefore c=100 \mathrm{~kJ} $$

A \(1500-\mathrm{kg}\) automobile traveling at \(30 \mathrm{~m} / \mathrm{s}\) is brought to rest by impacting a shock absorber composed of a piston with small holes that moves in a cylinder containing water. How much heat must be removed from the water to return it to its original temperature? As the piston moves through the water, work is done due to the force of impact moving with the piston. The work that is done is equal to the kinetic energy change; that is, $$ W=\frac{1}{2} m V^{2}=\left(\frac{1}{2}\right)(1500)(30)^{2}=675000 \mathrm{~J} $$ The first law for a cycle requires that this amount of heat must be transferred from the water to return it to its original temperature; hence, \(Q=675 \mathrm{~kJ}\).

4.29 Refrigerant R134a enters the compressor of the simple refrigeration cycle of Fig. \(4.19\) as a saturated vapor at \(120 \mathrm{kPa}\), enters the condenser at \(800 \mathrm{kPa}\) and \(50^{\circ} \mathrm{C}\), exits the condenser as a saturated liquid, and exits the evaporator as a saturated vapor. Calculate \((a)\) the lowest cycle temperature, \((b)\) the heat transferred from the condenser, \((c)\) the cooling load \(\dot{Q}_{E},(d)\) the necessary compressor horsepower, and \((e)\) the coefficient of performance for this refrigerator. The refrigerant mass flow rate is \(0.2 \mathrm{~kg} / \mathrm{s}\). The enthalpies are of primary interest into and out of the four control volumes. The IRC Fluid Property Calculator will be used to obtain the enthalpies, but, if, for some reason, it is not available, Appendix D can be used (the IRC Fluid Property Calculator gives slightly different values than does Table D). They are listed in the following: State 1 (saturated vapor): \(\quad P_{1}=P_{4}=120 \mathrm{kPa}, x=1 . \therefore h_{1}=237 \mathrm{~kJ} / \mathrm{kg}\) State 2 (superheated vapor): \(\quad P_{2}=800 \mathrm{kPa}, T_{2}=50^{\circ} \mathrm{C} . \therefore h_{2}=287 \mathrm{~kJ} / \mathrm{kg}\) State 3 (saturated liquid): \(\quad P_{3}=P_{2}=800 \mathrm{kPa}, x_{3}=0 . \therefore h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) State 4 (quality region): \(\quad P_{4}=P_{1}=120 \mathrm{kPa}, h_{4}=h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) Using the above enthalpies, the quantities of interest can be determined. (a) The lowest cycle temperature is \(T_{4}\) (refer to Fig. 4.19). The IRC Fluid Property Calculator, with \(P_{4}=120 \mathrm{kPa}\) and \(h_{4}=95.5 \mathrm{~kJ} / \mathrm{kg}\), provides \(T_{4}=-22.3^{\circ} \mathrm{C}\). (b) The heat transferred from the condenser (a heat exchanger) is $$ \dot{Q}_{C}=\dot{m}\left(h_{2}-h_{3}\right)=0.2 \mathrm{~kg} / \mathrm{s} \times(287-95.5) \text { But } / \mathrm{lbm}=38.3 \mathrm{~kJ} / \mathrm{s} $$ (c) The cooling load provided by the evaporator (a heat exchanger) is $$ \dot{Q}_{E}=\dot{m}\left(h_{1}-h_{4}\right)=0.2(237-95.5)=28.3 \mathrm{~kJ} / \mathrm{s} $$ (d) The compressor work requirement is $$ \dot{W}_{\text {Comp }}=\dot{m}\left(h_{2}-h_{1}\right)=0.2 \times(287-237)=10 \mathrm{~kW} \quad \text { or } \quad 13.4 \mathrm{hp} $$ (e) The coefficient of performance for this ideal cycle is $$ \mathrm{COP}=\frac{\text { desired effect }}{\text { required input }}=\frac{\dot{Q}_{E}}{\dot{W}_{\text {Comp }}}=\frac{28.3}{10}=2.83 $$ If the desired effect was to heat a space, \(\dot{Q}_{C}\) would be used as the numerator in COP.

Calculate the change in enthalpy of air which is heated from \(300 \mathrm{~K}\) to \(700 \mathrm{~K}\) if (a) \(C_{p}=1.006 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\). (b) Table B.5 is used. (c) The gas tables are used. (d) Compare the calculations of \((a)\) and \((b)\) with \((c)\). SOLUTION (a) Assuming the constant specific heat, we find that $$ \Delta h=C_{p}\left(T_{2}-T_{1}\right)=(1.006)(700-300)=402 \mathrm{~kJ} / \mathrm{kg} $$ (b) If \(C_{p}\) depends on temperature, we must integrate as follows: $$ \Delta h=\frac{1}{M} \int_{T_{1}}^{T_{2}} C_{p} d T=\frac{1}{28.97} \int_{300}^{700}\left(28.11+0.00197 T+0.48 \times 10^{-5} T^{2}-1.97 \times 10^{-9} T^{3}\right) d T=415 \mathrm{~kJ} / \mathrm{kg} $$ (c) Using Table E.1, we find \(\Delta h=h_{2}-h_{1}=713.27-300.19=413 \mathrm{~kJ} / \mathrm{kg}\). (d) The assumption of constant specific heat results in an error of \(-2.6\) percent; the expression for \(C_{p}\) produces an error of \(+0.48\) percent. All three methods are acceptable assuming the gas tables are most accurate.
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