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?

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VIII Problem Solving
19/10/2017

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?

The cold reservoir should have an absolute zero temperature.
The condition interprets to: efficiency =1-T_C/(T_H )=1-0/T_H =100%. The condition is an ideal situation which is impractical because the maximum efficiency for a Carnot engine is less than 100%

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?

QC= QH- W
W is the work done = 100J
QC is the heat lost in the process = 70J
The change in the internal energy is the heat absorbed by the gas QH
QH = QC + W
QH = (70 + 100) = 170J

An engine’s fuel is heated to 2,000 K and the surrounding air is 300 K. Calculate the ideal efficiency of the engine.

Carnot Efficiency is given by:
e_c= (T_H-T_C)/T_H
e_c=1-(T_c/T_H )
e_c=1-(300/2000)=0.85
e_c=85%

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.

e_c=1-(T_c/T_H )
Mr. White made an error by computing the efficiency using degrees Celsius instead of converting the temperatures to Kelvin. His wrong computation is shown below:
e_c=1-(10/100)=0.9=90%
From the measurements:
TC = (10 + 273) = 283K
TC = (100 + 273) = 373K
e_c=1-(283/373)
e_c=0.241
e_c=24.1%

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.
Q=mL
Q = (0.1 kg) (3.33 x 10^5 J/kg)
Q=335kJ

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.
Temperature difference is given by:
∆T= T_new- T_old
∆T= 〖98.6〗^0 F- 〖98.2〗^0 F
∆T=〖0.4〗^0 F
T(in Celsius)=5/9 (Farenheit-32)
In degree Celsius:
∆T=〖5/9(0.4〗^0-32)
∆T=-〖17.556〗^0 C

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?
NOT ANSWERED

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?
From first law of thermodynamics,
dQ=dW+dU
Therefore dQ=1000-3000= -2000 J
The student lost her heat of 2000J

A system does 100 J of work on its surroundings and gains 50 J of heat in the process. Calculate the change in the internal energy of the system and the surroundings.
dQ=dW+dU
Therefore dU=dQ-dW= 50-100= -50 J

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.

NOT ANSWERED

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?
e_c=1-(T_c/T_H )
e_c=1-(300/600)=0.5
e_c=50%
Efficiency η= Output/Input
Work done is the Output energy:
Output= η ×Input
Output= 0.5 ×3000
Output= 1500J

A heat engine has an efficiency of 40% and yields 5,000 J of work. What are the input and output heats?

Input Heat=W/η=5000/0.4=12500 J
Output Heat=Input Heat-Work
= 12500J-5000J
=7500 J