Describe thermodynamic concepts and their applications.
This assignment will allow to demonstrate the following objectives:
- Describe thermodynamic concepts and their applications.
- Extend the first law of thermodynamics to various daily life activities.
- Identify the maximum efficiency of a heat engine.
- Explain the role of latent heat while phases are changing.
Below is a detailed working of the questions choosen.
- 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:
Tc=0
Explanation:
The condition to have 100% efficiency in a carnot cycle is e=1-Tc/TH=1,
Thus Tc/TH = 0, As TH is different of zero we conclude that
Tc = 0 Kelvins.
- 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:
First law of thermodynamics = Total energy of an isolated system is constant.
U=Q-W
Work done = 100 J, Heat lost = -70 J
Therefore change in internal energy U=-70+100= 30 J
The change in the internal energy is 30J. The temperature of the gas raises and thus work is done to the system.
- 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: E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
Explanation:
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
E =1 – Tc/TH
where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.
Note: The unit of temperature must be in Kelvin.
Tc = 300K
TH = 2000K
Substituting the values of E, we have;
E = 1 – 300K/2000K
E = 1 – 0.15
E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
- 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:
The error he made was that he didn’t convert the unit of temperature to Kelvin.
The correct efficiency is 24%
Explanation:
Parameters given:
Temperature of hot reservoir = 100°C = 373 K
Temperature of cold reservoir = 10°C = 273 K
The efficiency of a heat engine is given as:
E = 1 – (Qc/QH) = 1 – (Tc/Th)
Where
Qc = Output heat;
QH = Input heat;
TC = Temperature of the cold reservoir;
TH = Temperature of the hot reservoir.
=> E = 1 – (283/373)
E = 1 – 0.76
E = 0.24
In percentage,
E = 0.24 * 100 = 24%
Hence, the efficiency of the engine is actually 24%.
The error he made was that he didn’t convert the temperature to Kelvin. If we leave the temperatures in °C, we have that:
E = 1 – (10/100)
E = 1 – 0.1 = 0.9
In percentage,
E = 0.9 * 100 = 90%
- 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:
33.5 kJ
Explanation:
In this case there is no change in temperature. The only thing that happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.
we have ice 100g =0.1 kg
Applying the equation Q=mL
Q= 0.1 kg * 335,000 J/kg
Q = 33,500 J
Q = 33.5 kJ
- 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:
I=98.6oF II=98.2oF
F=9/5C+32; C=5(F-32)*9
C=5(98.6-32)*9 = 37 oC
C=5(98.6-32)*9 = 36.7778 oC
Therefore the difference in degrees is 37oC – 36.7778oC = 0.2222 oC
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:
M = 0.5kg
Q = 2000J
C = 4,186 J/kg0C
dt=? Q= cmDt Dt= Q/cm Dt= 2000/ (0.5) (4,186) = 0.9 = 1oC
- 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:
Heat= Q
W = work = -1000 J
U= internal = -3000 J
Change U=Q-W, (-3000J)+ 1000J = -2000 J Conclusion: She loses heat.