Darryl English
2022-07-14
Answered

A rigid tank initially contains saturated vapor at 0,35 MPa. Then heat transferred to water until its temperature is ${400}^{\circ}C$. Determine the heat transfer during this process, in kJ/kg.

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bardalhg

Answered 2022-07-15
Author has **15** answers

Saturation Pressure: The pressure exerted by the vapor on the liquid when both liquid and vapor are in equilibrium is known as saturation pressure. the corresponding temperature is known as the saturation temperature.

First Law of thermodynamics: The first law of thermodynamics states that Energy can never be created nor be destroyed but it can be transferred from one form to another form.

First law of thermodynamics is given by formula

dQ = dU + W

dQ is the heat added or rejected

dU is the change in internal energy

W is the work done

Given data: The initial pressure of the saturated vapor ${P}_{1}=0.35\text{MPa}$

At 0.35 MPa from saturated water steam tables

The specific internal energy of saturated vapour is given by

${u}_{g}={u}_{1}=2548.5\text{kJ/kg}$

The saturated vapour is heated until the temperature reaches ${T}_{2}={400}^{\circ}C$

From superheated steam tables at 0.35 Mpa and ${400}^{\circ}C$

The specific internal energy of the vapour is given by

${u}_{2}=2965.4\text{kJ/kg}$

From the first law of thermodynamics

The heat transferred is given by

$Q=\mathrm{\u25b3}U+P\mathrm{\u25b3}V$

Here volume of the vessel is constant so $\mathrm{\u25b3}V=0$

$Q={u}_{2}-{u}_{1}\phantom{\rule{0ex}{0ex}}=2965.4-2548.5\phantom{\rule{0ex}{0ex}}=416.9\text{}\text{kJ/kg}$

First Law of thermodynamics: The first law of thermodynamics states that Energy can never be created nor be destroyed but it can be transferred from one form to another form.

First law of thermodynamics is given by formula

dQ = dU + W

dQ is the heat added or rejected

dU is the change in internal energy

W is the work done

Given data: The initial pressure of the saturated vapor ${P}_{1}=0.35\text{MPa}$

At 0.35 MPa from saturated water steam tables

The specific internal energy of saturated vapour is given by

${u}_{g}={u}_{1}=2548.5\text{kJ/kg}$

The saturated vapour is heated until the temperature reaches ${T}_{2}={400}^{\circ}C$

From superheated steam tables at 0.35 Mpa and ${400}^{\circ}C$

The specific internal energy of the vapour is given by

${u}_{2}=2965.4\text{kJ/kg}$

From the first law of thermodynamics

The heat transferred is given by

$Q=\mathrm{\u25b3}U+P\mathrm{\u25b3}V$

Here volume of the vessel is constant so $\mathrm{\u25b3}V=0$

$Q={u}_{2}-{u}_{1}\phantom{\rule{0ex}{0ex}}=2965.4-2548.5\phantom{\rule{0ex}{0ex}}=416.9\text{}\text{kJ/kg}$

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Bernoulli's principle is an example of which law of thermodynamics. Explain why?

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Suppose you mix two identical blocks of metal, one having a temperature of ${10}^{\circ}$ and the other ${20}^{\circ}$. Is it possible for the cold block to cool to ${5}^{\circ}C$ and the warm block to warm to ${25}^{\circ}C$?

A. Yes, because it adheres to the second law of thermodynamics.

B. No, because it will violate the zeroth law of thermodynamics.

C. Yes, because it adheres to the zeroth law of thermodynamics.

D. No, because it will violate the first law of thermodynamics.

A. Yes, because it adheres to the second law of thermodynamics.

B. No, because it will violate the zeroth law of thermodynamics.

C. Yes, because it adheres to the zeroth law of thermodynamics.

D. No, because it will violate the first law of thermodynamics.

asked 2022-09-03

I am trying to derive certain formulas and and I feel like the following expression is good for my derivation.

${\left(\frac{\mathrm{\partial}u}{\mathrm{\partial}T}\right)}_{p}={\left(\frac{\mathrm{\partial}u}{\mathrm{\partial}T}\right)}_{V}$

I am trying to say that the above expression is true because of first law of thermodynamics. any suggestions weather if this approach is good or not

${\left(\frac{\mathrm{\partial}u}{\mathrm{\partial}T}\right)}_{p}={\left(\frac{\mathrm{\partial}u}{\mathrm{\partial}T}\right)}_{V}$

I am trying to say that the above expression is true because of first law of thermodynamics. any suggestions weather if this approach is good or not

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Deriving $P{V}^{n}=\text{constant}$ for polytropic processes?

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The first law of thermodynamics

$\text{d}Q=\text{d}U+p\text{d}V-\mu \text{d}N$

is equivalent to saying: Q is a differentiable real valued function of U,V and N, so that we can write $\text{d}Q=\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}\text{d}U+\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}\text{d}V+\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}\text{d}N$, and $\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}=1$. Also, we define $p=\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}$ and $-\mu =\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}$

Is that it?

$\text{d}Q=\text{d}U+p\text{d}V-\mu \text{d}N$

is equivalent to saying: Q is a differentiable real valued function of U,V and N, so that we can write $\text{d}Q=\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}\text{d}U+\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}\text{d}V+\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}\text{d}N$, and $\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}=1$. Also, we define $p=\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}$ and $-\mu =\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}$

Is that it?

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A person weighs a fish of mass m = 40.0 N** **on a spring scale attached to the ceiling of an elevator. Show that if the elevator accelerates either upward or downward,

with a = 2.00 m/s2, the spring scale gives a reading that is different from the weight of the fish.

asked 2022-07-22

Change in state {for example from liquid to gas} is an isothermal process but the change in internal energy is not zero in this process isn't this contradictory?