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course Phy 242

2-27Had trouble on the last 2 questions

Ph2 110209

`q001. The same process we modeled in class last time has the following characteristics:

The temperatures of the hot and cold sources are T_c and T_h, initial volume is V_0, initial pressure is P_0, water density is rho and the acceleration of gravity is g.

The system is heated at constant volume, starting at T = T_c, until the temperature is T, where T_c < T < T_h, raising water in a thin tube whose end is open to the atmosphere.

The system is allowed to expand at constant pressure until it comes to final temperature T = T_h, displacing water as it expands. The water exits from the end of the tube.

The system is then cooled to its original temperature and pressure, without doing additional work or absorbing more thermal energy from the hot source.

In terms of T_c, T_h, V_0, P_0, rho and g:

How much PE does the system gain in the process?

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`dPE = ((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g

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Assuming the gas to be diatomic, how much thermal energy flows into the system during the process?

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State#1

`dQ = C_v*n*(T_i-T_c) = (5/2)*R*((P_0*V_0)/(R*T_c))*(T_i-T_c)

Because during this stage specific heat is at constant volume

Stage#2

`dQ = C_p*n*(T_i-T_c) = (7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

Because during this stage specific heat is at constant pressure

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What therefore is the ratio of PE gain to thermal energy absorbed by the system?

****

`dPE_gain/`dQ_in

[((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g]/

[(5/2)*R*((P_0*V_0)/(R*T_c))*(T_i-T_c) +(7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)]

= [((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g]/

(5/2)*R*((P_0*V_0)/(R*T_c))[(T_i-T_c + (7/5) (T_h-T_i)]

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How would your answer differ if the gas was monatomic, and why?

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Instead of having

(7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

We have

(5/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

For the thermal energy absorbed during the process when molar specific heat at constant Pressure

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`q002. By heating the air in a bottle from 273 K to 300 K, you have raised water in a thin vertical tube to a height of about 100 cm.

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By how much will the water level in the tube change, per Kelvin, if the temperature is changed?

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`dT = 27K

`dy = 100cm

100cm/27K = approx. 3.7 cm/K

!!!!!I did this incorrectly, after looking at notes I’m going back to make necessary corrections!!!!!!!!!!

Pressure at 300k

1atm*(`dT) = 1atm*(300/273) = `dP

Because starting pressure was

For T = 273, P = 1atm

Now knowing what the change in pressure is wrt `dT we can calculate `dy

rho*g*`dy = `dP

`dy = `dP/rho*g

Now we can say

-->Finding `dP for specific `dT then we can relate that to a specific `dy and find change in water level wrt change in temp.

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Assume the tube has cross-sectional area 0.1 cm^2. This could make a small difference in your preceding answer, which was based on the assumption that the volume of the tube is negligible. We won't worry about that for the preceding, but we will need to consider it for the following.

However the cross-sectional area of the tube makes a difference to the following:

If the tube makes a sharp right-angle turn at the 100 cm height (becoming horizontal), then how many cm of the additional tube will be filled, per Kelvin, if the temperature increases?

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For P_0 = atm, P_1 = 1atm(300/273) = approx. 1.1atm,T_0 = 273K, T_2 = 300K

Column of water supported at 1.1atm is

has volume of

c.s. area*ht = 0.1 cm^2*100cm = 10cm^3

has mas of

1gr/cm^3*10cm^3 = 10gr

has weight of

m*g = .01kg*9.8m/s^2 = .098N or approx. .1N

has force/area of

force/area = .098N/.1cm^2 = .098N/(.1 * 10^-4m^2) =

= 9800N/m^2

Now for `dT = 1K, we know that the pressure will change

P_2 = P_1*(301/300) = approx. 1.1037atm or approx. 110,370Pa

???I’m lost in my thinking. I think that ????

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Would the our results change if the air in the bottle had initially been at 290 Kelvin (assume that the water level and the position at which the tube makes its right-angle turn are adjusted accordingly)?

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If the other positions were changed as well there might be a difference but I pretty sure there would be. This entire system is all related to proportionalities of one variable to the other. With P_0 = 1atm the pressure change will not be as great with`dT = 10K

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What would change in our analysis if instead of a right-angle turn, the angle changed by a little less than 90 degrees, so that the tube has an upward slope of 0.1?

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the would gain more PE so that it would take more KE to move the water through the tube

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`q003. A wall is constructed of two different materials of equal thickness, the first with double the thermal conductivity of the other. The inner and outer walls are maintained at a constant temperature difference until a steady flow of thermal energy through the wall is achieved.

At that time, which wall will have the greater temperature gradient, and what will be the ratio of the temperature gradients? Hint: In the steady state, equal amounts of thermal energy pass through both materials.

!!!!!!!!I accidentally set this up backwards where temp grad #2 has twice the thermal conductivity!!!!!!!!!!

1st heat current.

k_1(`dT_1/`dx_1), where

k_1 is constant for thermal conductivity dependent on material

`dT_1 = change in temp over the interval of the material

`dx_1 is thickness of that part of the wall

2nd heat current.

2*k_1(`dT_1/`dx_1), where all symbols are the same except has twice the

thermal conductivity

I stated info backwards but 2nd heat current should have the greater temp grad, because the energy transfer are equal and 1st heat current has 2 times better thermal conductivity.

1st temp grad. is (1/2) the temp grad of 2nd temp grad.

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What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?

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temp grad = (1/2)(`dT_2/`dx_2), we know `dx is the same for both walls so `dT would (1/2)`dT_2.

Ratio = `dT/2`dx

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If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?

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mid pt of 10Celsius and 30Celsius is 20Celsius.

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If the first wall is also 50% thick than the other, how will the steady-state temperature gradients compare?

****

For `

dT_1/(2`dx_1), for an increase in thickness of the wall from 1 to 1.5 we have

`dT_1/(3`dx_1)

This could also be written as 2nd temp grad will be 3 times greater than the first.

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What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?

For

2*k_1(`dT_1/`dx_1) = k_2*3*(`dT_2/`dx_2)

1st layer has 2/3 the temp change of the 2nd

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If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?

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Approx 23.3333Celsius

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`q004. (University Physics) The PE gain of the system analyzed in the previous class is V0 ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * rho g y.

If the system is heated from temperature T_c to temperature T_h, at constant volume, what will be the maximum pressure and what therefore will be the maximum possible value of y?

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P_0(T_h/T_c) - P_0 = P_0((T_h/T_c) - 1)

y = P_0((T_h/T_c) - 1)/rho*g

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Substitute this maximum possible value of y for into the equation and simplify. What is your result, and why does this result make sense?

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Because PE_gain is equal to water*dist travled*g

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If the system is heated from temperature T_c to temperature T_h, at the constant initial pressure P_0, then how much water will be displaced, and to what height y? How much PE change will therefore occur?

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V_0( (T_h/T_c) - 1)

y = P_0( (T_i/T_c) - 1)

`dPE = V_0( (T_h/T_i) - 1) * P_0/rho*g( (T_i/T_c) - 1) *g

Which is

= Volume of water raised*ht raised to*accel of gravity

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We can factor all constant factors from the rest of the expression, obtaining the form

`dPE = rho g V0 * ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * y.

Find the value of y at which this expression is maximized.

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Find the value of `dPE for this maximizing value of y.

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Sketch a graph of `dPE vs. y.

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`q005. The system first analyzed in the preceding class had a rectangular P vs. V graph, due to the way the system was first cooled until it reached the original pressure.

The third phase of the cycle changes if we allow the gas to expand suddenly. Instead of a vertical line segment we get a part of the adiabatic curve P V^gamma = constant. This results in a different initial temperature T_3 and a different initial volume V_3 for the fourth phase of the cycle (that phase is still isobaric, just over a different range of volumes).

What will be the expressions for T_3 and V_3?

****

We would use system of equations where we know P_3 = P_0

PV^(gamma) is constant

???????I’m lost on this question and the following, if you could give me some pointers I can resubmit but I have no idea where to go????????????????

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The PE change of the system remains the same as before. However the work done by the system does differ from our previous result? What is the expression for the work done by this system?

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How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?

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there is less time so there would be less transfer

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`q006. The third phase of the cycle could also be accomplished isothermally, adding thermal energy to the (more slowly) expanding gas until it reaches the original temperature. Since T and n remain constant, the graph will in this case follow an isothermal curve P V = constant. Again T_3 and V_3 will differ from the values obtained for previous cycles, as will the amount of energy absorbed from the hot sink and the amount absorbed by the cold sink.

What will be the expressions for T_3 and V_3?

****

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What is the expression for the work done by this system?

****

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How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?

****

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How does the amount of thermal energy absorbed from the hot sink differ between the original cycle and this one?

****

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Self-critique (if necessary):

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Self-critique rating:

________________________________________

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02-27-2011

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course Phy 242

2-27Had trouble on the last 2 questions

Ph2 110209

`q001. The same process we modeled in class last time has the following characteristics:

The temperatures of the hot and cold sources are T_c and T_h, initial volume is V_0, initial pressure is P_0, water density is rho and the acceleration of gravity is g.

The system is heated at constant volume, starting at T = T_c, until the temperature is T, where T_c < T < T_h, raising water in a thin tube whose end is open to the atmosphere.

The system is allowed to expand at constant pressure until it comes to final temperature T = T_h, displacing water as it expands. The water exits from the end of the tube.

The system is then cooled to its original temperature and pressure, without doing additional work or absorbing more thermal energy from the hot source.

In terms of T_c, T_h, V_0, P_0, rho and g:

How much PE does the system gain in the process?

****

`dPE = ((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g

#$&*

Assuming the gas to be diatomic, how much thermal energy flows into the system during the process?

****

State#1

`dQ = C_v*n*(T_i-T_c) = (5/2)*R*((P_0*V_0)/(R*T_c))*(T_i-T_c)

Because during this stage specific heat is at constant volume

Stage#2

`dQ = C_p*n*(T_i-T_c) = (7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

Because during this stage specific heat is at constant pressure

#$&*

What therefore is the ratio of PE gain to thermal energy absorbed by the system?

****

`dPE_gain/`dQ_in

[((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g]/

[(5/2)*R*((P_0*V_0)/(R*T_c))*(T_i-T_c) +(7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)]

= [((P_0*(T_i/T_C) - P_0)/rho*g)*(V_0*(T_h/T_i) - V_0)*rho*g]/

(5/2)*R*((P_0*V_0)/(R*T_c))[(T_i-T_c + (7/5) (T_h-T_i)]

#$&*

How would your answer differ if the gas was monatomic, and why?

****

Instead of having

(7/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

We have

(5/2)*R*((P_0*V_0)/(R*T_c))*(T_h-T_i)

For the thermal energy absorbed during the process when molar specific heat at constant Pressure

#$&*

`q002. By heating the air in a bottle from 273 K to 300 K, you have raised water in a thin vertical tube to a height of about 100 cm.

****

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By how much will the water level in the tube change, per Kelvin, if the temperature is changed?

****

`dT = 27K

`dy = 100cm

100cm/27K = approx. 3.7 cm/K

!!!!!I did this incorrectly, after looking at notes I’m going back to make necessary corrections!!!!!!!!!!

Pressure at 300k

1atm*(`dT) = 1atm*(300/273) = `dP

Because starting pressure was

For T = 273, P = 1atm

Now knowing what the change in pressure is wrt `dT we can calculate `dy

rho*g*`dy = `dP

`dy = `dP/rho*g

Now we can say

-->Finding `dP for specific `dT then we can relate that to a specific `dy and find change in water level wrt change in temp.

#$&*

However the cross-sectional area of the tube makes a difference to the following:

If the tube makes a sharp right-angle turn at the 100 cm height (becoming horizontal), then how many cm of the additional tube will be filled, per Kelvin, if the temperature increases?

****

For P_0 = atm, P_1 = 1atm(300/273) = approx. 1.1atm,T_0 = 273K, T_2 = 300K

Column of water supported at 1.1atm is

has volume of

c.s. area*ht = 0.1 cm^2*100cm = 10cm^3

has mas of

1gr/cm^3*10cm^3 = 10gr

has weight of

m*g = .01kg*9.8m/s^2 = .098N or approx. .1N

has force/area of

force/area = .098N/.1cm^2 = .098N/(.1 * 10^-4m^2) =

= 9800N/m^2

Now for `dT = 1K, we know that the pressure will change

P_2 = P_1*(301/300) = approx. 1.1037atm or approx. 110,370Pa

???I’m lost in my thinking. I think that ????

#$&*

****

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What would change in our analysis if instead of a right-angle turn, the angle changed by a little less than 90 degrees, so that the tube has an upward slope of 0.1?

****

the would gain more PE so that it would take more KE to move the water through the tube

#$&*

!!!!!!!!I accidentally set this up backwards where temp grad #2 has twice the thermal conductivity!!!!!!!!!!

1st heat current.

k_1(`dT_1/`dx_1), where

k_1 is constant for thermal conductivity dependent on material

`dT_1 = change in temp over the interval of the material

`dx_1 is thickness of that part of the wall

2nd heat current.

2*k_1(`dT_1/`dx_1), where all symbols are the same except has twice the

thermal conductivity

I stated info backwards but 2nd heat current should have the greater temp grad, because the energy transfer are equal and 1st heat current has 2 times better thermal conductivity.

1st temp grad. is (1/2) the temp grad of 2nd temp grad.

#$&*

What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?

****

temp grad = (1/2)(`dT_2/`dx_2), we know `dx is the same for both walls so `dT would (1/2)`dT_2.

Ratio = `dT/2`dx

#$&*

If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?

****

mid pt of 10Celsius and 30Celsius is 20Celsius.

#$&*

If the first wall is also 50% thick than the other, how will the steady-state temperature gradients compare?

****

For `

dT_1/(2`dx_1), for an increase in thickness of the wall from 1 to 1.5 we have

`dT_1/(3`dx_1)

This could also be written as 2nd temp grad will be 3 times greater than the first.

#$&*

What therefore is the ratio of the temperature change through the first layer, to the temperature change through the second?

For

2*k_1(`dT_1/`dx_1) = k_2*3*(`dT_2/`dx_2)

1st layer has 2/3 the temp change of the 2nd

#$&*

If the inner wall is at temperature 30 Celsius and the outer at temperature 10 Celsius, what is the temperature at the point where the layers meet?

****

Approx 23.3333Celsius

#$&*

`q004. (University Physics) The PE gain of the system analyzed in the previous class is V0 ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * rho g y.

If the system is heated from temperature T_c to temperature T_h, at constant volume, what will be the maximum pressure and what therefore will be the maximum possible value of y?

****

P_0(T_h/T_c) - P_0 = P_0((T_h/T_c) - 1)

y = P_0((T_h/T_c) - 1)/rho*g

#$&*

Substitute this maximum possible value of y for into the equation and simplify. What is your result, and why does this result make sense?

****

Because PE_gain is equal to water*dist travled*g

#$&*

****

V_0( (T_h/T_c) - 1)

y = P_0( (T_i/T_c) - 1)

`dPE = V_0( (T_h/T_i) - 1) * P_0/rho*g( (T_i/T_c) - 1) *g

Which is

= Volume of water raised*ht raised to*accel of gravity

#$&*

We can factor all constant factors from the rest of the expression, obtaining the form

`dPE = rho g V0 * ( (T_h / T_c) * ( P_0 / (P_0 + rho g y) ) - 1) ) * y.

Find the value of y at which this expression is maximized.

****

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Find the value of `dPE for this maximizing value of y.

****

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Sketch a graph of `dPE vs. y.

****

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`q005. The system first analyzed in the preceding class had a rectangular P vs. V graph, due to the way the system was first cooled until it reached the original pressure.

What will be the expressions for T_3 and V_3?

****

We would use system of equations where we know P_3 = P_0

PV^(gamma) is constant

???????I’m lost on this question and the following, if you could give me some pointers I can resubmit but I have no idea where to go????????????????

#$&*

The PE change of the system remains the same as before. However the work done by the system does differ from our previous result? What is the expression for the work done by this system?

****

#$&*

How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?

****

there is less time so there would be less transfer

#$&*

What will be the expressions for T_3 and V_3?

****

#$&*

What is the expression for the work done by this system?

****

#$&*

How does the amount of thermal energy transferred to the cold sink differ between the original cycle and this one?

****

#$&*

How does the amount of thermal energy absorbed from the hot sink differ between the original cycle and this one?

****

#$&*

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

02-27-2011