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Chapter 4: Problem 1

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, \(Q=675 \mathrm{~kJ}\).

Short Answer

Expert verified
675 kJ of heat must be removed.

Step by step solution

01

Determine the Kinetic Energy

Calculate the initial kinetic energy of the automobile using the formula for kinetic energy: Step 1: \(\frac{1}{2} m v^{2}\).
02

Insert the Given Values

Substitute the mass (1500- kg ) and velocity (30 m/s ) into the kinetic energy formula: \(KE = \left(\frac{1}{2}\right)(1500)(30)^\right) = 675000 J\).
03

Recognize the Relationship Between Work and Heat

Note that the work done to bring the car to rest (675000 J ) converts to heat that must be removed from the water to restore its original temperature. According to the first law of thermodynamics, this amount of work is also the amount of heat (Q ) that must be transferred.
04

Convert the Units from Joules to Kilojoules

Convert the result from Joules to Kilojoules by dividing by 1000: \(675000 J = 675 kJ\).
05

Final Answer

The heat that must be removed from the water to return it to its original temperature is 675 kJ.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. The kinetic energy of an object depends on its mass and velocity. The formula to calculate kinetic energy is given by \(KE = \frac{1}{2} m v^2\).
When you plug in the mass (m) of the automobile as 1500 kg and its velocity (v) as 30 m/s, you get \(KE = \frac{1}{2} (1500) (30)^2 = 675000 \mathrm{J}\).
This means the automobile initially has 675000 Joules of kinetic energy.
This energy can be transformed or transferred to different forms, according to physical laws like the First Law of Thermodynamics.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another or transferred between systems.
In the context of the exercise, the kinetic energy of a moving car is converted into work as the car comes to rest. This work is then transferred to the water in the shock absorber in the form of heat.
Mathematically, the First Law can be represented as \(\Delta U = Q - W\), where:
  • \Delta U\ is the change in internal energy
  • \Q\ is the heat added to the system
  • \W\ is the work done by the system
Since the car is brought to rest, all the kinetic energy is used up in doing work, so the heat Q transferred equals the work done (675000 J, or 675 kJ).
Work-Energy Principle
The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy.
In simpler terms, if you apply a force to move an object, the energy exerted by the force (work) will be the same as the increase or decrease in the object's kinetic energy.
For the automobile, initially in motion, work must be done to bring it to rest, which translates to removing its kinetic energy.
Using the relation \(W = \Delta KE\), we see that the work done to stop the car (W) is equal to the initial kinetic energy the car had: 675000 J.
This means work converts the entire kinetic energy into other forms, mostly heat in this scenario.
Heat Removal
When the kinetic energy of the moving automobile is converted into work by the shock absorber, this energy is transferred as heat to the water in the shock absorber.
To restore the water to its original temperature, this heat must be removed.
The amount of heat removed (Q) is the same as the work done, which equals the initial kinetic energy, 675000 J or 675 kJ.
This process of removing the heat ensures the water returns to its starting temperature, maintaining thermal equilibrium.
Heat removal is a practical application of the First Law of Thermodynamics, demonstrating energy conservation and transfer.

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Most popular questions from this chapter

A turbine accepts superheated steam at 800 psia and \(1200^{\circ} \mathrm{F}\) and rejects it as saturated vapor at 2 psia (Fig. 4.26). Predict the horsepower output if the mass flux is \(1000 \mathrm{lbm} / \mathrm{min}\). Also, calculate the velocity at the exit. Assuming zero heat transfer, the energy equation (4.66) provides us with $$ -\dot{W}_{T}=\dot{m}\left(h_{2}-h_{1}\right)=\left(\frac{1000}{60}\right)(1116.1-1623.8)=-8462 \mathrm{Btu} / \mathrm{s} \quad \text { or } \quad 11970 \mathrm{hp} $$ where Tables C.3E and C.2E have provided the enthalpies. By (4.58), using \(\rho=1 / v\) $$ V_{2}=\frac{v \dot{m}}{A}=\frac{(173.75)(1000 / 60)}{\pi(2)^{2}}=230 \mathrm{ft} / \mathrm{s} $$

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.

Water enters a radiator through a 4-cm-diameter hose at \(0.02 \mathrm{~kg} / \mathrm{s}\). It travels down through all the rectangular passageways on its way to the water pump. The passageways are each \(10 \times 1 \mathrm{~mm}\) and there are 800 of them in a cross section. How long does it take water to traverse from the top the bottom of the 60 -cm-high radiator? SOLUTION The average velocity through the passageways is found from the continuity equation, using \(\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}\) : $$ \dot{m}=\rho_{1} V_{1} A_{1}=\rho_{2} V_{2} A_{2} \quad \therefore V_{2}=\frac{\dot{m}}{\rho_{2} A_{2}}=\frac{0.02}{(1000)[(800)(0.01)(0.001)]}=0.0025 \mathrm{~m} / \mathrm{s} $$ The time to travel \(60 \mathrm{~cm}\) at this constant velocity is $$ t=\frac{L}{V}=\frac{0.60}{0.0025}=240 \mathrm{~s} \quad \text { or } \quad 4 \mathrm{~min} $$

The pressure of \(200 \mathrm{~kg} / \mathrm{s}\) of water is to be increased by \(4 \mathrm{MPa}\). The water enters through a 20 -cm-diameter pipe and exits through a \(12-\mathrm{cm}\)-diameter pipe. Calculate the minimum horsepower required to operate the pump. SOLUTION The energy equation (4.68) provides us with $$ -\dot{W}_{p}=\dot{m}\left(\frac{\Delta P}{\rho}+\frac{V_{2}^{2}-V_{1}^{2}}{2}\right) $$ The inlet and exit velocities are calculated as follows: $$ V_{1}=\frac{\dot{m}}{\rho A_{1}}=\frac{200}{(1000)(\pi)(0.1)^{2}}=6.366 \mathrm{~m} / \mathrm{s} \quad V_{2}=\frac{\dot{m}}{\rho A_{2}}=\frac{200}{(1000)(\pi)(0.06)^{2}}=17.68 \mathrm{~m} / \mathrm{s} $$ The energy equation then gives $$ \dot{W}_{P}=-200\left[\frac{4000000}{1000}+\frac{(17.68)^{2}-(6.366)^{2}}{2}\right]=-827200 \mathrm{~W} \quad \text { or } \quad 1109 \mathrm{hp} $$ Note: The above power calculation provides a minimum since we have neglected any internal energy increase. Also, the kinetic energy change represents only a 3 percent effect on \(\dot{W}_{P}\) and could be neglected.

Estimate the constant-pressure specific heat and the constant-volume specific heat for R134a at 30 psia and \(100^{\circ} \mathrm{F}\). SOLUTION We write the derivatives in finite-difference form and, using values on either side of \(100^{\circ} \mathrm{F}\) for greatest accuracy, we find $$ \begin{aligned} &C_{p} \equiv \frac{\Delta h}{\Delta T}=\frac{126.39-117.63}{120-80}=0.219 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \\ &C_{v} \equiv \frac{\Delta u}{\Delta T}=\frac{115.47-107.59}{120-80}=0.197 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \end{aligned} $$
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