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__**Week 2:**__
 * 1.** A log-raft was made by lashing six logs together, where each log has a diameter 0.25 m and length 1.90 m. How many children can safely afloat on the raft in sea water, if the average weight of a child is 200 N? Density of the logs is 800 kg/m3 and sea water is 1024 kg/m3.

To find the volume of a log, use the formula V=(pi)(r^2)(height) Multiply this number by 6 to get the total volume of all the logs. Find the Buoyant force: buoyant force=(volume of logs)(density of sea water)(gravity) To find the weight of the logs you must multiply their density, volume and gravity. The total buoyant force is equal to the weight of the logs + the weight of the children. Subtract the weight of the logs from the buoyant force to get the total weight of all of the children. Divide this number by 200 (the weight of 1 child) to get the total amount of children the raft can hold.


 * 2.**A cube of wood whose edge is 0.015 m is floating in a liquid (in a glass beaker) with one of its faces parallel to the liquid surface. The density of wood is 550 kg /m³, that of liquid is 1240 kg/m³. How far below the liquid surface is the bottom face of the cube?

First, you need to find the volume of the cube. To find the weight of the wood, use the formula, mg=(volume of the wood)(density of the wood)(gravity) To find the buoyant force multiply (density of liquid)(gravity)(area of object)(height of object) The buoyant force is equal to the weight of the object: weight of wood=(density of liquid)(gravity)(area of object)(height of object) From here, you can solve for the height of the object.


 * Professor's Note: Very well done, Sarah, Keep up the good work !! **

__**Week 3:**__ **1.**  An automobile having a mass of 1000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring of constant 5 × 10^6 N/m and compresses 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost during impact with the wall? You are given the following: m=1000kg k=5 × 10^6 N/m x=3.16 cm

Because this problem states that energy is conserved, you can use the equation 1/2mv^2=1/2kx^2. Using this equation, you can solve for the velocity of the car (after you have converted x to meters).

**2**. A block of unknown mass is attached to a spring of spring constant 6 .5 N/m and undergoes simple harmonic motion with an amplitude of 10 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 30 cm/s.

i. Calculate the mass of the block.

Use the conservation of energy equation 1/2mv^2=1/2kx^2 and plug in all of the given variables. Then solve for m.

ii. Find the period of the motion.

You can solve for the period using the following equation: T=2pi*sqrt(m/k)

iii.Calculate the maximum acceleration of the block.

The sum of the forces acting on a block is F=ma. Using hook's law you also know that F=-kx Therefore you know that ma=-kx You can plug in all of the given values and solve for the acceleration from here.


 * please refer to the professor's page here .... Professor **

**3.** A 0.4 kg block attached to a spring of force constant 12 N/m oscillates with an amplitude of 8 cm. i. Find the maximum speed of the block.

You can solve for the maximum speed using the following equation: max velocity=(amplitude)(angular frequency) You also know that angular frequency=sqrt(k/m) Then you can substitute the values for the angular frequency and amplitude into the first equation to solve for the maximum velocity.

ii. Find the speed of the block when it is 4 cm from the equilibrium position. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">x(t)=4cm=.04m <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">x(t)=(amplitude)( angular frequency)sin((angular frequency)(time)) You can plug in all of the given values and solve for time. Plug the value of t into the equation v(t)=(amplitude)(angular frequency)(sin((angular frequency)(time)) From here you can solve for the velocity.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">iii. Find its acceleration at 4 cm from the equilibrium position. To solve this problem, you can use the following equation: a(t)=-(angular frequency)^2(x(t)) From here you can solve for the acceleration.


 * Professor's Note : Sarah, once again great work !! **

01. A transverse wave of frequency 35 Hz propagates down a string. Two points 5 cm apart are out of phase by π/3 i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !)
 * __ Week 4: __**

Convert cm to meters. Solve for the wavelength (lambda) using this equation: distance/phase=lambda/period

ii. At a given point, what is the phase difference (as a multiple of π ) between two ‘displacements for times 5 ms apart?

First you must solve for the period: T=1/f Use the following equation to solve for the displacement: x/t=2pi/T Because you already know the values for t and T, you can solve for x.

02. At t = 0, a transverse wave pulse in a wire is described by the function y(x,t)=0=6/(x^2+3) where x and y are in meters. The pulse is traveling in the positive x direction with a speed of 4.5 m/s. What is y at time = t? (Hint : find y(x,t) )

Set the derivative of the position equal to the velocity: dx/dt=4.5 Integrate the left side from x1 to x2 and integrate the right side from 0 to t. Find x2 in terms of x1 and t. Substitute into original equation for position.

__ Week 5: __ 1. A light string of mass per unit length 8 g/m has its ends tied to two walls separated by a distance equal to three fourths the length L of the string. A mass m is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9.8 m/s^2.What size mass should be suspended from the string to produce a wave speed of 60 m/s?

First, you must draw a force diagram. Next use Newton's 1st law and set the sum of the forces in the y-direction equal to 0. Using this, you will find that 2Tsin(theta)=mg From here you will find that m=(2Tsin(theta))/g Use the equation v=sqrt(T/mu) Rearrange this equation to solve for T: T=(mu)v^2 Substitute this into the previous equation: m=(2(mu)v^2sin(theta))/g

2. A simple pendulum consists of a ball of mass M = 5 kg hanging from a uniform string of mass m = 0.6 g and unknown length L. The pendulum is located in gravitational field g = 9.8 m/s² and has period T = 2 s. Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. (For simplicity, neglect the string’s weight compared to the ball’s weight. Likewise, neglect the ball’s radius compared to the string’s length). Answer in units of m/s.

Normally, the tension is equal to 2pi*sqrt(L/g) But because it is at rest F=mg You can substitute this into the equation v=sqrt(F/(mu)) to get v=sqrt(mg/(mu))

__**<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">Week 6: **__ **<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">No Quiz **

__**<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">Week 7: **__

<span style="font-family: DejaVu Sans Light,sans-serif;">1.The “red shift” of radiation from a distant galaxy consists of the light known to have a wavelength of 434 nm when observed in the laboratory, appearing to have a wavelength of 462 nm. What is the speed of galaxy in the line of sight relative to the Earth? The speed of light is 2.99792 × 10^8 m/s.

> observed frequency=frequency of source*(1/(1+(observed velocity/velocity of source))
 * 1) <span style="font-family: DejaVu Sans Light,sans-serif;">Convert nanometers to meters.
 * 2) <span style="font-family: DejaVu Sans Light,sans-serif;">Find the observed frequency and the frequency of the source using the equation v=f*wavelength which can be rewritten as f=v/wavelength.
 * 3) <span style="font-family: DejaVu Sans Light,sans-serif;">You can find the observed velocity using the following equation:

<span style="font-family: DejaVu Sans Light,sans-serif;">2.The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves.

<span style="font-family: DejaVu Sans Light,sans-serif;">Beta=loudness of 1 man = 20 dB <span style="font-family: DejaVu Sans Light,sans-serif;">B1=20 dB > Beta4=loudness of 4 men > Beta4=10log(4I/Io) > Beta4=10log(4)+10log(I/Io) > Beta4=B1+10log(4)
 * 1) <span style="font-family: DejaVu Sans Light,sans-serif;">Use the formula Beta= 10log(I/Io)
 * 2) <span style="font-family: DejaVu Sans Light,sans-serif;">Multiply I/Io by 4 to find the loudness of 4 men.

Professor's Note : Very good answers.... good job !!

<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">__**Week 8**__

<span style="font-family: DejaVu Sans Light,sans-serif;">__Polarization__-the orientation of oscillations in the plane perpendicular to a transverse wave's direction of travel.

<span style="font-family: DejaVu Sans Light,sans-serif;">1. Initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of Ø1 = Ø2 = Ø3 = 50⁰ with the directions of the y axis. What percentage of the initial intensity is transmitted by the system? (hint: be careful with the angles)

<span style="font-family: DejaVu Sans Light,sans-serif;">Because the initial light is unpolarized when it is being transmitted through the first sheet, you must use the one-half rule: <span style="font-family: DejaVu Sans Light,sans-serif;">I1=Io/2 <span style="font-family: DejaVu Sans Light,sans-serif;">After the light has gone through the first sheet, it becomes polarized. Therefore. You must use the cosine-squared rule: <span style="font-family: DejaVu Sans Light,sans-serif;">I2=I1cos^2(50) <span style="font-family: DejaVu Sans Light,sans-serif;">I3=I2cos^2(50)

<span style="font-family: DejaVu Sans Light,sans-serif;">Now you substitute to find I3 in terms of Io: <span style="font-family: DejaVu Sans Light,sans-serif;">I2=(Io/2)cos^2(50) <span style="font-family: DejaVu Sans Light,sans-serif;">I3=(Io/2)cos^2(50)cos^2(50)

<span style="font-family: DejaVu Sans Light,sans-serif;">To find the percentage of the initial intensity that is being transmitted by the system, you must multiply the coefficient in front of Io by 100.

<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">__**Week 9:**__

<span style="font-family: DejaVu Sans Light,sans-serif;">1.A fruit fly of height H sits in front of a lens (say lens 1) on the central axis. The lens forms an image of the fly at a distance d = 20 cm from the fly ; the image has the same orientation as the fly and a height = 2 H.

<span style="color: #000000; font-family: DejaVu Sans Light,sans-serif;">a) What are the focal length of the lens and the object distance of the fly?

<span style="font-family: DejaVu Sans Light,sans-serif;">You know that the difference between the fly and the image is 20 cm. You can write an equation to express this: <span style="font-family: DejaVu Sans Light,sans-serif;">i-p=20

<span style="font-family: DejaVu Sans Light,sans-serif;">|m|=h'/h where h' is the image height and h is the object height. You are given that h'=2H and h=H: <span style="font-family: DejaVu Sans Light,sans-serif;">|m|=2H/H <span style="font-family: DejaVu Sans Light,sans-serif;">|m|=2

<span style="font-family: DejaVu Sans Light,sans-serif;">You also know that m=-i/p <span style="font-family: DejaVu Sans Light,sans-serif;">However, because you are given the absolute value of m, you must take the absolute value of -i/p <span style="font-family: DejaVu Sans Light,sans-serif;">m=i/p <span style="font-family: DejaVu Sans Light,sans-serif;">2=i/p <span style="font-family: DejaVu Sans Light,sans-serif;">Now you can express one as a function of the other: <span style="font-family: DejaVu Sans Light,sans-serif;">i=2p

<span style="font-family: DejaVu Sans Light,sans-serif;">Substitute this into the equation i - p =20 (GOOD POINT) <span style="font-family: DejaVu Sans Light,sans-serif;"> to solve for p. Then substitute the value you find for p in the equation i=2p to solve for I.

<span style="font-family: DejaVu Sans Light,sans-serif;">Because the orientation of the object and the image are the same, you know that the image formed is a virtual image. Therefore, I is negative.

<span style="font-family: DejaVu Sans Light,sans-serif;">Plug the values for I and p into the equation 1/f=1/i + 1/p and solve for the focal length.

<span style="color: #000000; font-family: DejaVu Sans Light,sans-serif;">Fly then leaves lens 1 and sits in front of a second lens (say lens 2), which also forms an image at d = 20 cm from the fly and has the same orientation as of the fly, but now the image height is 0.5 H

<span style="color: #000000; font-family: DejaVu Sans Light,sans-serif;">b) What are the focal length of the second lens and the new object distance?

<span style="font-family: DejaVu Sans Light,sans-serif;">Use the equation |m|=h'/h to find m given that h'=.5H and h=H.

<span style="font-family: DejaVu Sans Light,sans-serif;">M=.5 <span style="font-family: DejaVu Sans Light,sans-serif;">|m|=|-i/p| <span style="font-family: DejaVu Sans Light,sans-serif;">.5=i/p <span style="font-family: DejaVu Sans Light,sans-serif;">i=.5p

<span style="font-family: DejaVu Sans Light,sans-serif;">The difference between the object and the image is 20 cm: <span style="font-family: DejaVu Sans Light,sans-serif;">p-i = 20 --> (GOOD POINT) <span style="font-family: DejaVu Sans Light,sans-serif;">Substitute the equation for I from the last equation into this equation and solve for p. <span style="font-family: DejaVu Sans Light,sans-serif;">After you have solved for p, plug this value into the equation above and solve for I.

<span style="font-family: DejaVu Sans Light,sans-serif;">Because the orientation of the image and object are the same, a virtual image is formed. This means I is negative.

<span style="font-family: DejaVu Sans Light,sans-serif;">Use the equation 1/f=1/p+1/i to solve for the focal length. Great job Sarah !! You have got this right and so far you are the only one who has it right -- professor My name is Stacey, not Sarah. <span style="font-family: DejaVu Sans Light,sans-serif;">2. When the rectangular metal tank in the figure, is filled to the top with an unknown liquid, Observer O with eyes level with the top of the tank, can just see corner E. A ray that refracts toward O at the top surface of the liquid is shown. If D = 85.0 cm and L = 1.1 m, what is the index of refraction of the liquid?

<span style="font-family: DejaVu Sans Light,sans-serif;">Using trigonometry, you can solve for Ɵ1: <span style="font-family: DejaVu Sans Light,sans-serif;">Tan Ɵ1=L/D <span style="font-family: DejaVu Sans Light,sans-serif;">Ɵ1=arctan(L/D)

<span style="font-family: DejaVu Sans Light,sans-serif;">Use Snell's law to solve for the refractive index of the liquid: <span style="font-family: DejaVu Sans Light,sans-serif;">n1sinƟ1=n2sinƟ2

<span style="font-family: DejaVu Sans Light,sans-serif;">Because the observer's eyes are level with the top of the tank and the axis is normal to the liquid surface, Ɵ2=90. You know that n2=1 because the refractive index of air is 1.

<span style="font-family: DejaVu Sans Light,sans-serif;">Plug these values into the equation above and solve for n1.

<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">__**Week 10**__ : <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">a) In a double slit arrangement, the slits S1 and S2 are illuminated with light whose wavelength is 643 nm. The slit separation is 0.15 mm, slit to screen distance is 145 cm. What is the spacing between the bright fringes on the viewing screen?

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">Because you know that there are two slits and there are bright fringes, you must use the equation dsinƟ=mλ where sinƟ=y/L.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">You know that m=1. Therefore dy/L=λ. <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">You can rearrange this to solve for y: <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">y= λL/d

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">Plug in the given values and solve for y.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">b) A thin sheet of transparent plastic of thickness t = 0.48 μm with a refractive index of n = 1.5 is placed over only the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y, find y.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">For this problem, N1=N2. N1 is the # of wavelengths along r1 and N2 is the # of wavelengths along r2.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">dsinƟ=r2-r1 <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">If you rearrange this equation, you will find that r2=dsinƟ+r1.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">N2=r2/λ <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">When you substitute the equation for r2 above, you will end up with N2=(dsinƟ+r1)/λ

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">To find N1, you must add the # of wavelengths along r1 before the sheet to the # of wavelengths passing through the sheet.

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;"># of wavelengths along r1 before the sheet=(r1-t)/λ <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;"># of wavelengths along r1 passing through the sheet=t/λ_n where λ_n=λ/n_sheet <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">N1=(r1+t(n-1))/λ

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">N1=N2 <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">(dsinƟ+r1)/λ=(r1+t(n-1))/λ

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">This can be simplified: <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">dsinƟ=t(n-1) <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">sinƟ=(t(n-1))/d

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">y=LsinƟ <span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">y=L(t(n-1))/d

<span style="font-family: DejaVu Sans Light,sans-serif; font-size: 14.6667px;">__** Week 11 **__ <span style="font-family: DejaVu Sans Light,sans-serif;">1) A rectangular plate has an area of 95 cm². If the temperature increases by 112 ºC, calculate the increase in the area. Use 5 × 10-6 (ºC)-1 as an average coefficient of linear expansion, and ignore terms of second order.

<span style="font-family: DejaVu Sans Light,sans-serif;">To calculate the increase in area, you must use the following formula: <span style="font-family: DejaVu Sans Light,sans-serif;">ΔA=AβΔT

<span style="font-family: DejaVu Sans Light,sans-serif;">where: <span style="font-family: DejaVu Sans Light,sans-serif;">A=Area <span style="font-family: DejaVu Sans Light,sans-serif;">β=2α (α is the coefficient of linear expansion) <span style="font-family: DejaVu Sans Light,sans-serif;">ΔT= Temperature Change

<span style="font-family: DejaVu Sans Light,sans-serif;">2)A quantity of ideal gas at 10 ºC and 100 kPa occupies a volume of 2.50 m³ . (a). How many moles of the gas are present? If the pressure is now raised to 300 kPa and the temperature is raised to 30 ºC, how much volume does the gas occupy? Assume there are no leaks.

<span style="font-family: DejaVu Sans Light,sans-serif;">To calculate the moles of gas present, you must use the ideal gas law which states: <span style="font-family: DejaVu Sans Light,sans-serif;">pV=nRT

<span style="font-family: DejaVu Sans Light,sans-serif;">Rearrange this to solve for n: <span style="font-family: DejaVu Sans Light,sans-serif;">n=(pV)/(nRT) <span style="font-family: DejaVu Sans Light,sans-serif;">All of the values are given. Remember to convert the pressure to Pa and R is a constant.

<span style="font-family: DejaVu Sans Light,sans-serif;">You can also arrange the ideal gas law to solve for the volume: <span style="font-family: DejaVu Sans Light,sans-serif;">V=(nRT)/p

<span style="font-family: DejaVu Sans Light,sans-serif;">Plug in the given values and solve to find the numerical answer.

<span style="color: #800080; font-family: DejaVu Sans Light,sans-serif;">__**Weeks 14 & 15**__ <span style="font-family: DejaVu Sans Light,sans-serif;">1. Three moles of a certain diatomic molecular gas are heated at constant pressure from 300 K to 400 K. The heat capacity of this gas under constant pressure is 28.8 J/mol · K and the universal gas constant is 8.31451 J/mol · K. Calculate the heat transferred to the gas. Answer in units of J.

<span style="font-family: DejaVu Sans Light,sans-serif;">To calculate the heat that is transferred to the gas, you must use the formula Q=n(c_p)ΔT <span style="font-family: DejaVu Sans Light,sans-serif;">n=# of moles <span style="font-family: DejaVu Sans Light,sans-serif;">c_p=molar specific heat at constant pressure (Note: c_p=c_v+R) <span style="font-family: DejaVu Sans Light,sans-serif;">ΔT=Change in Temperature

<span style="font-family: DejaVu Sans Light,sans-serif;">From here, you can plug in all your given values and solve.

<span style="font-family: DejaVu Sans Light,sans-serif;">2. A vessel with a capacity of 5 L contains 0.125 mol of an ideal gas at 1.5 atm. The value of Boltzmann’s constant is 1.38066 × 10−23 J/K, and the universal gas constant is 8.31 J/K · mol. What is the average translational kinetic energy of a single molecule? Answer in units of J.

<span style="font-family: DejaVu Sans Light,sans-serif;">(KE)inst=1/2mv^2 <span style="font-family: DejaVu Sans Light,sans-serif;">(KE)avg=1/2m(v^2)avg <span style="font-family: DejaVu Sans Light,sans-serif;">(KE)avg=1/2m(v_rms)^2 <span style="font-family: DejaVu Sans Light,sans-serif;">v_rms=sqrt(3RT/M) <span style="font-family: DejaVu Sans Light,sans-serif;">(KE)avg=1/2m(3RT/M) <span style="font-family: DejaVu Sans Light,sans-serif;">m/M=N_A <span style="font-family: DejaVu Sans Light,sans-serif;">(KE)avg=(3RT/2N_A) <span style="font-family: DejaVu Sans Light,sans-serif;">k=R/N_A <span style="font-family: DejaVu Sans Light,sans-serif;">(KE)avg=3kT/2

<span style="font-family: DejaVu Sans Light,sans-serif;">k=Boltzmann's Constant <span style="font-family: DejaVu Sans Light,sans-serif;">R=Universal Gas Constant <span style="font-family: DejaVu Sans Light,sans-serif;">T=Temperature

<span style="font-family: DejaVu Sans Light,sans-serif;">3. A 50.0 g block of copper whose temperature is at 150 F is placed in an insulating box with a 100 g block of lead whose temperature is 100 F (a) What is the equilibrium temperature of the two-block system?

<span style="font-family: DejaVu Sans Light,sans-serif;">ΔQ=0 <span style="font-family: DejaVu Sans Light,sans-serif;">Q (absorbed) + Q (released) = 0 <span style="font-family: DejaVu Sans Light,sans-serif;">-(mcΔT)copper=(mcΔT)lead <span style="font-family: DejaVu Sans Light,sans-serif;">-m1c1(Tf-150)=m2c2(Tf-100)

<span style="font-family: DejaVu Sans Light,sans-serif;">m1 =mass of copper <span style="font-family: DejaVu Sans Light,sans-serif;">m2 = mass of lead <span style="font-family: DejaVu Sans Light,sans-serif;">c1 =specific heat of copper <span style="font-family: DejaVu Sans Light,sans-serif;">c2 = specific heat of lead

<span style="font-family: DejaVu Sans Light,sans-serif;">Plug in given values and solve for Tf.

<span style="font-family: DejaVu Sans Light,sans-serif;">(b) What is the change in the internal energy of the system?

<span style="font-family: DejaVu Sans Light,sans-serif;">This is an adiabatic process. The system is insulated and therefore no energy transfers occur between the system and its surrounding environment (Q=0).

<span style="font-family: DejaVu Sans Light,sans-serif;">(ΔE)int=Q-W <span style="font-family: DejaVu Sans Light,sans-serif;">(ΔE)int=-W

<span style="font-family: DejaVu Sans Light,sans-serif;">The system also has a constant volume-->W=0 <span style="font-family: DejaVu Sans Light,sans-serif;">(ΔE)int=0

<span style="font-family: DejaVu Sans Light,sans-serif;">(c) What is the change in the entropy of the system?

<span style="font-family: DejaVu Sans Light,sans-serif;">(ΔS)total= ΔS1+ΔS2 <span style="font-family: DejaVu Sans Light,sans-serif;">ΔS=Sf-Si <span style="font-family: DejaVu Sans Light,sans-serif;">ΔS=integral of (dQ/T) <span style="font-family: DejaVu Sans Light,sans-serif;">dQ=mc(dT) <span style="font-family: DejaVu Sans Light,sans-serif;">dT=Tf-Ti <span style="font-family: DejaVu Sans Light,sans-serif;">ΔS=mc*ln(Tf/Ti)

<span style="font-family: DejaVu Sans Light,sans-serif;">(ΔS)total=m1c1*ln(Tf/Ti)+m2c2*ln(Tf/Ti)

<span style="font-family: DejaVu Sans Light,sans-serif;">4. A 600 W carnot engine operates between constant temperature reservoirs at 100 C and 60 C.(a) What is the rate at which energy is taken in by engine as heat(b) What is the rate at which energy is exhausted by the energy as heat?

<span style="font-family: DejaVu Sans Light,sans-serif;">(a) First, you must figure out the engine's efficiency: <span style="font-family: DejaVu Sans Light,sans-serif;">Efficiency=1-(T_L/T_H) <span style="font-family: DejaVu Sans Light,sans-serif;">*Note: T_Land T_H must be in Kelvin

<span style="font-family: DejaVu Sans Light,sans-serif;">Efficiency=|W|/|Q_H| <span style="font-family: DejaVu Sans Light,sans-serif;">W=energy produced <span style="font-family: DejaVu Sans Light,sans-serif;">Q_H=energy input

<span style="font-family: DejaVu Sans Light,sans-serif;">Rearrange the previous equation to get this: <span style="font-family: DejaVu Sans Light,sans-serif;">Q_H=W/efficiency

<span style="font-family: DejaVu Sans Light,sans-serif;">This will give you the input in W. 1 W=1 J/s.

<span style="font-family: DejaVu Sans Light,sans-serif;">(b) heat exhausted by engine=heat input – heat output <span style="font-family: DejaVu Sans Light,sans-serif;">heat exhausted by engine=answer from part a-600