Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,829 users

Physics Help?

0 votes
1)
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1*10^6 J of electricity with the hot reservoir at 500 K during Day One and then produces 1*10^6 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was: 
a. greatest on Day One. b. greatest on Day Two. c. the same on both days. d. zero on both days.

1.1 )
One kilogram of water at 1.00 atm at the boiling point of 100°C is heated until all the water vaporizes. What is its change in entropy? (For water, Lv = 2.26*10^6 J/kg) 
a. 12 100 J/K b. 6 060 J/K c. 3 030 J/K d. 1 220 J/K
2 )
What is the change in entropy (delta S) when one mole of silver (108 g) is completely melted at 961°C? (The heat of fusion of silver is 8.82 *10^4 J/kg.) 
a. 5.53 J/K b. 7.72 J/K c. 9.91 J/K d. 12.10 J/K.


3)
A cylinder containing an ideal gas has a volume of 2.0 m3 and a pressure of 1.0 *10^5 Pa at a temperature of 300 K. The cylinder is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. The change in internal energy of the gas is 6.0 *10^5 J. Find the change in entropy of the block associated with the heat transfer to the gas. 
a. 0
b. +670 J/K c. - 440 J/K d.-1 100 J/K 

 
 
4)
The surface of the Sun is at approximately 5 700 K and the temperature of the Earth's surface is about 290 K. What total entropy change occurs when 1 000 J of heat energy is transferred from the Sun to the Earth? 
a. 2.89 J/K b. 3.27 J/K c. 3.62 J/K d. 3.97 J/K
5)
1 On an average diet, the consumption of 10 liters of oxygen releases how much energy? (4.8 kcal are released per liter of oxygen consumed.) 
a. 48 kJ
b. 200 kJ c. 4.2 kJ
d. 4 200 kJ
6)
A person consumes 2 500 kcal/day while expending 3 500 kcal/day. In a month’s time, about how much weight would this person lose if the loss were essentially all from body fat? (Body fat has an energy content of about 4 100 kcal per pound.) 
a. 1 pound b. 2 pounds c. 7 pounds d. 15 pounds
7)
A pound of body fat has an energy content of about 4 100 kcal. If a 1 400-kg automobile had an equivalent amount of translational kinetic energy, how fast would it be moving? (0.447 m/s = 1 mph, 1 kcal = 4 186 J) 
a. 3.1 mph b. 14 mph c. 75 mph d. 350 mph.
 
 
8)
Suppose the ends of a 20-m-long steel beam are rigidly clamped at 0 C to prevent expansion. The rail has a cross-sectional area of 30 cm2. What force does the beam exert when it is heated to 40C? (alpha steel = 1.1 *10^5/C, Ysteel = 2.0 * 10^11 N/m2).
a. 2.6 * 105 N
 
9)

Estimate the volume of a helium-filled balloon at STP if it is to lift a payload of 500 kg. The density of air is 1.29 kg/m3 and helium has a density of 0.178 kg/m3.
a. 4 410 m3 b. 932 m3 c. 450 m3 d. 225 m3
 
10)

 Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C. The specific heat of the solid: 
a. is equal to that of the fluid. b. is less than that of the fluid. c. is more than that of the fluid.

11) A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 540 cal/g) 
a. +1.8 C° b. –1.8 C° c. +0.18 C° d. - 0.18 C°
100 g of liquid nitrogen at its boiling point of 77 K is stirred into a beaker containing 500 g of 15°C water. If the nitrogen leaves the solution as soon as it turns to gas, how much water freezes? The heat of vaporization of nitrogen is 48 cal/g and that of water is 80 cal/g.
a. none b. 29 g c. 68 g d. 109 g
Ans: None How do you solve? How would you calculate how much water froze if this weren't the case?

 

asked Nov 7, 2014 in PHYSICS by mthskey12 Rookie
edited Nov 7, 2014 by bradely
Provide complete question .

Sorry don't know how all of them were cut off. Replying with all the questions. 

Suppose the ends of a 20-m-long steel beam are rigidly clamped at 0

Sorry don't know how all of them were cut off. Replying with all the questions. Suppose the ends of a 20-m-long steel beam are rigidly clamped at 0

Can't get alny of my questions to show up. Trying a new thread sorry.
Replying with links to questions.

12 Answers

0 votes

(1)

The Temperature of the cold reservoir = 300 K.

The Temperature of the Hot reservoir on Day 1 = 500 K.

Then Efficiency of the Carnot cycle is

η = 1 - (Tc / Th)

η = 1 - (300 / 500)

η = 0.4 * 100

η = 40 %

The power plant produces 1 * 10^6 J of Energy.

We know that

η = (W / Qh)

0.4 = (1 * 10^6 / Qh)

Qh = (1 * 10^6 / 0.4)

Qh = 2500000 J

Qh = 2.5 * 10^6 J

Therefore heat put into the system Qh = 2.5 * 10^6 J

We know that heat put into the system = Heat Entering the system + Heat absorbed by the environment.

Qh = W + Qc

Qc = Qh - W

Qc = 2.5 * 10^6 - 1 * 10^6

Qc = 1.5 * 10^6 J.

Thermal Pollution is the amount of heat absorbed by the Environment.

Greater the the amount of heat absorbed by the Environment, Greater is the thermal pollution.

Thermal Pollution on Day 1 is the heat absorbed by the Environment is 1.5 * 10^6 J.

The Temperature of the cold reservoir = 300 K.

The Temperature of the Hot reservoir on Day 2 = 600 K.

Then Efficiency of the Carnot cycle is

η = 1 - (Tc / Th)

η = 1 - (300 / 600)

η = 0.5 * 100

η = 50 %

The power plant produces 1 * 10^6 J of Energy.

We know that

η = (W / Qh)

0.4 = (1 * 10^6 / Qh)

Qh = (1 * 10^6 / 0.5)

Qh = 2000000 J

Qh = 2.0 * 10^6 J

Therefore heat put into the system Qh = 2.0 * 10^6 J

We know that heat put into the system = Heat Entering the system + Heat absorbed by the environment.

Qh = W + Qc

Qc = Qh - W

Qc = 2.0 * 10^6 - 1 * 10^6

Qc = 1.0 * 10^6 J.

Thermal Pollution is the amount of heat absorbed by the Environment.

Greater the the amount of heat absorbed by the Environment, Greater is the thermal pollution.

Thermal Pollution on Day 2 is the heat absorbed by the Environment is 1.0 * 10^6 J.

Therefore the thermal pollution was greater on Day 1.

Therefore (a) is the correct option.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(1.1)

Mass of the water is 1 kg.

Temperature = 100°C = 100 + 273 = 373 K.

Specific heat of water vapourization is 2.26 * 10^6 J/kg

Heat of vapourisation Qw = mass * Specific Heat

Heat of vapourisation Qw = 1 * 2.26 * 10^6

Heat of vapourisation Qw = 2.26 * 10^6 J.

Change in Entropy = transfer of heat into that system / Thermodymanic Temperature.

ΔS = ΔQ / T

ΔS = (2.26 * 10^6) / 373

ΔS = 6058.98 J/K

Therefore change in Entropy is 6058.98 J/K.

Option (b) is the correct option.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(3)

The volume of the gas V = 2.0 m³

Pressure of the Gas P = 1.0 * 10^5 Pa.

Temperature T = 300 K

We know that Ideal gas Equation

PV = nRT

Where R is gas constant = 8.314 J/(K*mol)

PV = nRT

1*10^5 * 2.0 = n * 8.314 * 300

n = 80.1  moles

Now the cylinder is placed in the metal block.

Final Temperature of gas in cylinder = 900 K.

Pressure of the Gas P = 1.0 * 10^5 Pa.

n = 80.1 moles

We know that Ideal gas Equation

PV = nRT

Where R is gas constant = 8.314 J/(K*mol)

PV = nRT

(10^5) * V = 80.1 * 8.314 * 900

V = 5.99 m³

Volume of the gas at Temperature 900 K is 6 m³

Therefore,

Change in the Internal Energy at Constant Pressure

ΔH=ΔU+PΔV

Read more: http://www.physicsforums.com

ΔH=ΔU+PΔV

Where ΔH is Enthalpy = TΔS

T is the Temperature of the block.

ΔS is Entropy

ΔU is Change in the Internal Energy = 6 * 10^5

P is Pressure

ΔV is change in volume.

T * ΔS  = 6 * 10^5 + (10^5 *( 6 - 2))

900 * ΔS = 6 * 10^5 + (10^5 *4)

ΔS = 10 * 10^5 J / 900

ΔS = 1111.1 J / K

Therefore the change in the Entropy is 1111.1 J/K.

Option (d) is the correct answer.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(4)

The Temperature of the Sun = 5700 K

The Temperature of the Earth = 290 K

Heat Transferred = 1000 J.

Change in Entropy of the sun ΔSsun = ΔQ / T

Change in Entropy of the sun ΔSsun = 1000 / 5700

Change in Entropy of the sun ΔSsun = 0.1754 J/K

Change in Entropy of the Earth ΔSearth = ΔQ / T

Change in Entropy of the Earth ΔSearth =1000 / 290

Change in Entropy of the Earth ΔSearth = 3.448 J/K

Total Change in Entropy = Change in Entropy of the sun ΔSsun + Change in Entropy of the Earth ΔSearth

Change in Total Entropy = 0.1754 + 3.448 = 3.6234 J/K

Therefore the Total Entropy is 3.62 J/K.

Option (C) is the correct answer.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(5)

4.8 Kcal of energy is released per liter of Oxygen.

Amount of Energy Released per 10 liters of oxygen = 4.8 * 10

Amount of Energy Released per 10 liters of oxygen = 48 Kcal.

We know that 1 Kcal = 4.18 kJ

Amount of Energy Released per 10 liters of oxygen = 48  * 4.18 kJ.

Amount of Energy Released per 10 liters of oxygen = 200.64 kJ.

Therefore Amount of Energy Released per 10 liters of oxygen is 200.64 kJ.

Option (B) is the correct answer.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(2)

Mass of the silver is 108 g = 0.108kg.

Temperature = 961°C = 961 + 273 = 1234 K.

Latent heat of water vapourization is 8.82 * 10^4 J/kg

Heat of fusion Qw = mass * Latent Heat

Heat of fusion Qw = 0.108 * 8.82 * 10^4

Heat of fusion Qw = 9525.6 J.

Change in Entropy = transfer of heat into that system / Thermodymanic Temperature.

ΔS = ΔQ / T

ΔS = (9525.6.) / 1234

ΔS = 7.719 J/K

Therefore change in Entropy is 7.719 J/K.

Option (b) is the correct option.

answered Nov 7, 2014 by Lucy Mentor
edited Nov 7, 2014 by Lucy
0 votes

(6)

Consuming energy Ec = 2500 kcal/day

Expending energy Ee = 3 500 kcal/day

Body fat energy content Eb=  4 100 kcal per pound

Actual loss of energy from body fat Ea = Ee - Ec

Ea = 3500 - 2500

Ea = 1000 kcal/day

Actual loss of energy from body fat for 1 month Eam = 1000*30

Eam = 30000 kcal

Total weight loss for one month W = Ea/Eb

W = 30000/4100

W = 7.3

W ≈ 7 pounds

Weight loss of that person from his body fat is 7 pounds.

answered Nov 7, 2014 by Shalom Scholar
0 votes

(8)

Length of the steel Beam = 20 m

Young's Modulus = 2.0 * 10^11 N/m².

Cross sectional area of the Beam = 30 cm² = 30 * 10^-4 m

linear coefficient of thermal expansion of steel α = 13*10^-6 /K

We know that Change in the length Δl = Original Length L1 * linear coefficient of thermal expansion α  * Change in Temperature ΔT

Δl = l * α * ΔT 

Δl = 20 * 13*10^-6 * (40 - 0)

Δl = 0.0104 m.

Therefore the Increase in the length of the beam is 0.0104 m.

Young's modulus of steel = stress / strain

Where Stress = Force / Area

Strain = Change in Length / Original Length

Strain = 0.0104 m / 20 m = 5.2 * 10^-4.

Young's modulus of steel = stress / strain

Stress = (Young's modulus of steel) * (Strain)

Force / Area = (2.0 * 10^11) * ( 5.2 * 10^-4)

Force / Area = 104 * 10^6

Force = (104 * 10^6) * (30 * 10^-4)

Force = 312000 N.

Force = 312 kN

Therefore the force exerted by the beam is 312 kN.

answered Nov 7, 2014 by Lucy Mentor
0 votes

(7)

Body fat energy E = 4100 kcal

E = 4100*4186 = 17162600 J

Automobile mass m = 1400 kg

Kinetic energy E = ½mv²

17162600 = ½(1400)v²

v² = 17162600*2/1400

v² = 24518

v = 156.58 m/s

0.447 m/s = 1 mph ⇒ 1 m/s = (1/0.447) mph

v = 156.58 (1/0.447) mph

v = 350.29 mph

v ≈ 350 mph

Automobile moving velocity is 350 mph

answered Nov 7, 2014 by Shalom Scholar
0 votes

(10)

Solid temperature Ts = 70°C

Solid mass ms= 20 g

Solid specific heat = Cs

Fluid temperature Tf = 20°C

Fluid mass mf = 100 g

Fluid specific heat = Cf

Thermal equilibrium  temperature Te = 30°C

ΔTs = Te - Ts  = 70 - 30 = 40

ΔTf = Te - Tf  = 30 - 20 = 10

At thermal equilibrium

mfCf (ΔTf) = msCs(ΔTs)

100*Cf*10 = 20*Cs*40

1000Cf = 800Cs

Cs = 1.25 Cf

Cs > Cf

The specific heat of the solid is more than that of the fluid.

Answer is Option (c)

answered Nov 7, 2014 by Shalom Scholar

Related questions

asked Jun 29, 2017 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Nov 29, 2015 in PHYSICS by anonymous
...