The Role of Latent Heat While Phases Are Changing Paper Example

Instructions: Choose 8 of the 10 problems below. Show your work in detail. Answer the questions directly in this template. Before doing this, it is highly recommending that you thoroughly review the four examples in the Unit Lesson.

Question #1

The efficiency of a Carnot engine is e=1-Tc/TH, where Tc is a temperature of the cold reservoir and TH is a temperature of the hot reservoir. What is the condition to have 100% efficiency? Hint: What is the mathematical condition for Tc/TH to be zero.

Answer #1

Carnot engine efficiency is given by e=1-Tc/TH. For the efficiency to hit 100%, Tc/TH, must be 0. Mathematical application will require the cold reservoir temperature, Tc to be 0K, absolute zero. The condition is that heat must be rejected to sink at a temperature of 0 Kelvin. This achievement is only ideal but cannot be practically possible. However, the physical inference is that the only way to get all the heat transfer into doing work is to eliminate all the heat energy, and this needs the cold reservoir to be at absolute zero, 0K.

Question #2

Suppose the work done to compress a gas is 100 J. If 70 J of heat is lost in the process, what is the change in the internal energy of the gas? Hint: Use the first law of thermodynamics. The internal energy of a system changes due to heat (Q) and work (W): U=Q-W. The change in internal energy is equivalent to the difference between the heat added to the system and the work done by the system. Think if the work done is to the system or by the system. This determines the sign of W.

Answer #2

From the first law of thermodynamics, the change in internal energy, U, of a system is equal to the net heat transfer, Q, into the system plus the net work done, W, by or on the system. The equation of the system is U = Q-W. The sign of W is positive if work is done on the system and negative if work is done by the system. For this case, work is done on the system to compress the gas, therefore, a positive W. Heat is lost in the process, making the symbol of Q to be negative. Therefore, the change in the internal energy of the gas is as shown below:

U = Q-W

= -70+100

= 30 Joules

Question #3

An engine's fuel is heated to 2,000 K and the surrounding air is 300 K. Calculate the ideal efficiency of the engine. Hint: The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH : e=W/QH. W=QH - QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

Answer #3

The ideal efficiency, e, of a Carnot engine is given by the ratio of work done, W, by the engine divide by the input heat energy, QH. That is, e = W/QH. Work done, W is given by (QH - Qc). Therefore, e = (QH - Qc)/QH or e = 1 - Qc/QH. According to the French Engineer Carnot, the ratio Qc/QH is same as the ratio of temperatures Tc/TH. Therefore, Carnot Efficiency is also given by the equation e = (1 - Tc/TH) x 100. The value is converted into percentage by multiplying the result by 100%.

e = (1 - Tc/TH) x 100

e = (1 - 300/2000) x 100

e = (1 - 0.15) x 100

e = 85%

The ideal efficiency of the system is, therefore, 85%.

Question #4

Mr. White claims that he invented a heat engine with a maximum efficiency of 90%. He measured the temperature of the hot reservoir as 100o C and that of cold reservoir as 10o C. Find the error that he made and calculate the correct efficiency. Hint: The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH : e=W/QH. W=QH - QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

Answer #4

The efficiency of a Carnot's heat engine depends on both high and low temperatures. It is important to convert the temperatures into Kelvin before carrying out the efficiency calculation. This is because engines can operate even at a cold temperature of 0 degrees Celsius and below. The mistake Mr. White made in inventing his heat engine was to fail to convert the temperatures from degrees Celsius to Kelvin. His calculation that produced a mathematical efficiency of 90% was as shown below.

e = (1 - Tc/TH) x 100

e = (1 - 10/100) x 100

e = (1 - 0.1) x 100

e = 90%

The correct efficiency of the engine can be calculated after conversions of the reservoir cold and hot temperatures to Kelvin. That is;

e = (1 - Tc/TH) x 100

e = (1 - ((273+10)/(273+100)) x 100

e = (1 - 283/373) x 100

e = (1-0.759) x 100

e = 24.1%

The correct efficiency of the engine Mr. White invented is 24.1%.

Question #5

How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: The heat (Q) used to change from one phase to another phase of the matter is Q=mL, where L is the latent heat. Its unit is J/kg.

Answer #5

The amount of energy, Q, to melt ice into water is given by the multiplication of mass of the ice, m, with the latent heat of fusion, L. Q = mL. 100g into Kg is 100/1000 = 0.1 Kg. Q = 0.1 Kg X 335,000 J/Kg. Therefore, the energy is to convert ice into water at 0 degrees Celsius is33500 J or 33.5 KJ.

Question #6

It was determined in the 19th century that the normal human body temperature is 98.6o F. A more recent study found that it is 98.2o F. Express the difference in the temperature in Celsius. Hint: Use the converting formula between Fahrenheit and Celsius scales: F=9/5C +32. Be careful about the unit.

Answer #6

The difference in temperatures of the human body as determined in the 19th Century and in the 21st Century is given by:

(98.6o F- 98.2o F) = 0.4o F

Converting this temperature from Fahrenheit to degrees Celsius will utilize the formula:

F = (9/5)C + 32

Converting 98.6 o F to o C

C = 98.6 X (5/9) - 32

22.8 oCConverting 98.2 o F to o C

C = 98.2 X (5/9) - 32

22.6 oCThe difference in terms of Celsius is, therefore, 22.8 - 22.6 = 0.2 oC.

Question #7

Suppose 0.5 kg of blood flows from the interior to the surface of John's body while he is exercising. The released energy is 2,000 J. The specific heat capacity of blood is 4,186 J/kgo C. What is the temperature difference between when the blood arrives at the body surface and returns back to the interior of the body? Hint: Use the formula regarding heat Q, specific heat capacity c, mass m, and temperature change dT. Q= cm dT. Please look at p.290 in our textbook. Also, review Example 1 with its solution in Study Guide.

Answer #7

The formula to determine the quantity of energy is given by; Q = mC(dt), that is mass of blood by specific heat capacity of blood by the change in temperature. The quantity of heat released is 2000. The temperature difference is given by the formula, dt = Q/mC = 2000J/0.5kgX4186J/Kg oC. Therefore, dt =0.955 oC.Question #8

A student does 1,000 J of work when she moves to her dormitory. Her internal energy is decreased by 3,000 J. Determine the heat during this process. Does she gain or lose her heat? Hint: Use the first law of thermodynamics. The internal energy of a system changes due to heat Q and work done W: U=Q-W. Also, look at a similar case, Example 3 with its solution in Study Guide.

Answer #8

The first law of thermodynamics is given by U = Q-W. Internal energy decrease = -3000J and work done by the system is - 1000J. Therefore, the heat during the process is given as follows:

-3000J = Q - 1000J

Q = -3000 + 1000 J

Q = -2000 J.

The student loses heat as she walks to the dormitory.

Question #9

In a construction site, 2 kg of aluminum shows the increment of temperature by 5oC. Ignoring the work, what is the change in the internal energy of the material? The specific heat capacity of aluminum is 900 J/kg oC. Hint: The internal energy of a system changes due to heat Q and work done W: U=Q-W. If we ignore, the work, the internal energy U is identical to the heat Q of the system. We know that relation between heat Q, specific heat capacity c, mass m, and temperature change dT; Q= cm dT. That is, U=Q=cm dT.

Answer #9

Ignoring work, W, U = Q = mC(dt), where U = internal energy, m = mass of aluminum, and dt = change in temperature. U = Q = 2kg X 900J/Kg oC X 5 oC = 9000 J.

Question #10

The input heat of a Carnot engine is 3,000 J. The temperature of a hot reservoir is 600 K and that of a cold reservoir is 300 K. What is the work done? Hint: The efficiency e of a Carnot engine is defined as the ratio of the work done, W, by the engine to the input heat QH : e=W/QH. W=QH - QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

Use the formula, e=1-Tc/TH. Please review the Example 4 with its solution in Study Guide. Once you evaluate, you can find the work done of the system using the formula, e=W/QH

Answer #10

e = 1 - Tc/TH = 1 - 300/600

= (1 - 0.5)100 = 50%

e is also given as; e = W/QH;

W = e X QH.

W = (50/100) X 3000J

W = 1500 J