Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L
Answers
By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.
How to determine the boundary work done by a gas during an expansion process
A process is a consecution of states of a system. The boundary work (W), in kilojoules, is the work done by the system on surroundings and in a P-Vdiagram this kind of work is equal to the area below the curve, which can be approximated by Riemann sums:
(1)
Where:
- p - Pressure, in kilopascals.
- V - Volume, in cubic meters.
Now we proceed to calculate the boundary work:
W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]
W = 0.243 kJ
By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.
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Answer:
attached below
Explanation:
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Answers
Answer:
Explanation:
The specific heat for watermelon above freezing point is . The heat liberated by the watermelon to cool down to 8°C is:
The heat absorbed by the household refrigerator is:
Time needed to cool the watermelons is:
b. A rigid bar does not bend regardless of the loads acting upon it.
c. A rigid bar deforms when experiencing applied loads.
d. A rigid bar is unable to translate or rotate about a support.
e. A rigid bar represents an object that does not experience deformation of any kind.
Answers
Answer:
option b and E are true
Explanation:
A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:
option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.
option b is true. A rigid bar remains unchanged regardless of the load that it carries.
option c is incorrect, a rigid bar does not deform with loads on it
option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.
option e is true. A rigid bar would not experience any deformation whatsoever.
True
False
Answers
Answer: true
Explanation:
Answers
Answer:
Ig =7.2 +j9.599
Explanation: Check the attachment
B. Determine the overall heat transfer coefficients, U; and U., for the pipe.
C. Would your answer change if the two materials were swapped so that the inside material were carbon steel and the outside material of the pipe were made of AISI 304 stainless steel? If so, calculate new values for UA and q. If not, explain why your answer would not change. (Here, assume the dimensions of the inner material (now plain carbon steel) match those of the AISI 304 schedule 30 stainless steel from part A, and the stainless steel is 0.075 inches thick on the outside.)
Answers
Sorry man i dont know the answer to this one
Answers
Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
where,
E(t) is the modulus of elasticity at any temperature 'T'
is the modulus of elasticity at absolute zero.
is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.