T and V changing in an isobaric process

In summary: OUNT OF ENERGY THAT GOES INTO HEATING THE LAYER:We can calculate the mass of a unit column since the layer is in hydrostatic balance. The mass of the column of air is then m/A. The energy that is used to heat the layer is then ##p\Delta h##. This energy is used to convert thermal energy (radiation) into kinetic energy.
  • #1
BearY
53
8

Homework Statement


Consider a 1000 to 700 hPa layer in the atmosphere that is at rest and is in hydrostatic balance. The layer is subjected to radiative cooling at a rate of 5.0e3 J/s/m2 for one hour.
a) Calculate the change in mean temperature for the layer assuming all energy goes into heating the layer. Hint: We derived the 1st law of thermodynamics on a per mass basis.
b) Find the resulting decrease in the thickness of the layer.

Homework Equations



##\Delta Q = \Delta U + W ##
##\Delta p = -mg##
##\Delta T = \frac{\Delta U}{mC_p}##

The Attempt at a Solution


I am a bit confused and have a few questions.
What does "all energy goes into heating the layer" mean? From what I see, the energy goes in the layer is work done to the layer. Which is ##p\Delta h##
Also, is radiative cooling ##\Delta Q##, or ##\Delta U## only?
We can calculate the mass of a unit column since the layer is in hydrostatic balance.
Then the next step, from what I see, would depend on my understanding of the question.
Did I completely misunderstand the question?
 
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  • #2
BearY said:

Homework Statement


Consider a 1000 to 700 hPa layer in the atmosphere that is at rest and is in hydrostatic balance. The layer is subjected to radiative cooling at a rate of 5.0e3 J/s/m2 for one hour.
a) Calculate the change in mean temperature for the layer assuming all energy goes into heating the layer. Hint: We derived the 1st law of thermodynamics on a per mass basis.
b) Find the resulting decrease in the thickness of the layer.

Homework Equations



##\Delta Q = \Delta U + W ##
##\Delta p = -mg##
Be careful with units. LS units are force/area. RS units are force. Δp = -(m/A)g where m/A is the mass per unit area of a column of air in the layer.
##\Delta T = \frac{\Delta U}{mC_p}##

The Attempt at a Solution


I am a bit confused and have a few questions.
What does "all energy goes into heating the layer" mean? From what I see, the energy goes in the layer is work done to the layer. Which is ##p\Delta h##
Also, is radiative cooling ##\Delta Q##, or ##\Delta U## only?

We can calculate the mass of a unit column since the layer is in hydrostatic balance.
Then the next step, from what I see, would depend on my understanding of the question.
Did I completely misunderstand the question?
The problem is not stated all that clearly. It appears to me that the radiative cooling relates to heat flow in the form of thermal radiation from the Earth surface. You are to assume that all this heat flow is captured by the layer (ie. Q = 5000 Js-1m-2)

AM
 

Related to T and V changing in an isobaric process

What is an isobaric process?

An isobaric process is a thermodynamic process in which the pressure of the system remains constant while other variables, such as temperature and volume, can change.

How does temperature change in an isobaric process?

In an isobaric process, the temperature of the system can change as heat is added or removed. If heat is added, the temperature will increase, and if heat is removed, the temperature will decrease. This is due to the change in the internal energy of the system.

What is the relationship between temperature and volume in an isobaric process?

In an isobaric process, the volume of the system can change as temperature changes. This is known as the Charles' Law, which states that the volume of a gas is directly proportional to its temperature, assuming constant pressure.

How does T and V change in an isobaric process compared to an isochoric process?

In an isochoric process, the volume of the system remains constant, so there is no change in volume. However, in an isobaric process, the volume can change as temperature changes. This means that T and V change differently in these two processes.

Why is the T and V changing in an isobaric process important?

The change in temperature and volume in an isobaric process is important because it affects the work and heat transfer of the system, and therefore, the efficiency and performance of the process. It is also essential to understand in order to accurately predict and control the behavior of the system.

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