Thermal Physics- Can you make good tea at a certain temperature

[H.fulls]

Homework Statement

According to experts good tea can only be made at temperatures greater than 96 degrees celsius. If this is true, can you brew good tea at elevation 4.5km (Pressure = 6.2*10^{4}Pa) . Given that latent heat of vaporisation for water is 2.4*10^{6} J/kg and water has a molar mass of 18g.

Homework Equations

Clausius-clapeyron equation \frac{dp}{dT} = \frac{L}{T(V2-V1)}
but V1 is negligable so \frac{dp}{dT} = \frac{L}{TV2}

Ideal gas equation pV=nRT

The Attempt at a Solution

Tried to integrate so have \int\frac{1}{p}dp=\frac{L}{R}\int\frac{1}{T^{2}}dT
from substituting in V = \frac{RT}{p} from ideal gas equation.
But don't really know what to do from here! I don't know what limits to put in for the integral or if I am just going about this all wrong from the start!
Ive been told the clausius-clapeyron statement must be used by my professor!
Any suggestions would be helpful! Thankyou!

 

[gneill]

Homework Statement

According to experts good tea can only be made at temperatures greater than 96 degrees celsius. If this is true, can you brew good tea at elevation 4.5km (Pressure = 6.2*10^{4}Pa) . Given that latent heat of vaporisation for water is 2.4*10^{6} J/kg and water has a molar mass of 18g.

Homework Equations

Clausius-clapeyron equation \frac{dp}{dT} = \frac{L}{T(V2-V1)}
but V1 is negligable so \frac{dp}{dT} = \frac{L}{TV2}

Ideal gas equation pV=nRT

The Attempt at a Solution

Tried to integrate so have \int\frac{1}{p}dp=\frac{L}{R}\int\frac{1}{T^{2}}dT
from substituting in V = \frac{RT}{p} from ideal gas equation.
But don't really know what to do from here! I don't know what limits to put in for the integral or if I am just going about this all wrong from the start!
Ive been told the clausius-clapeyron statement must be used by my professor!
Any suggestions would be helpful! Thankyou!

Hi H.fulls, Welcome to Physics Forums.

You might be interested having a look at the following arrangement of the clausius-clapeyron relationship:
$$ ln\left(\frac{P_1}{P_2}\right) = \frac{H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
Where the P's are the vapor pressures and the T's the temperatures (in K).

 

[H.fulls]

Thanks! :)

I was getting this integral before but not sure what to do with it! And is H_{vap} the same as my latent heat value? I have no idea what values to use for P_{1}, P_{2}, T_{1} or T_{2} :S

 

[gneill]

Thanks! :)

I was getting this integral before but not sure what to do with it! And is H_{vap} the same as my latent heat value? I have no idea what values to use for P_{1}, P_{2}, T_{1} or T_{2} :S

Yes, H_vap is the latent heat of vaporization. The P's are the vapor pressures at temperatures T₁ and T₂. What do you know about the relationship between ambient atmospheric pressure and vapor pressure of a fluid at its boiling point?

 

[H.fulls]

Ah okay :) Erm..nothing probably! But I seem to recall that the ambient pressure and pressure at boiling point are the same? maybe... ? haha :)

 

[gneill]

Ah okay :) Erm..nothing probably! But I seem to recall that the ambient pressure and pressure at boiling point are the same? maybe... ? haha :)

Yup.

 

[H.fulls]

But if the pressures are the same does that not make the left hand side ln(1) =0! ?

 

[gneill]

But if the pressures are the same does that not make the left hand side ln(1) =0! ?

Why do say that the pressures are the same? The atmospheric pressure is different for the two cases.

 

[H.fulls]

Ah okay so the P's denote just the atmospheric pressure? I know the pressure for not boiling, but I don't know the pressure for when it is boiling! So confused by this question!

 

[gneill]

Ah okay so the P's denote just the atmospheric pressure? I know the pressure for not boiling, but I don't know the pressure for when it is boiling! So confused by this question!

The P's represent vapor pressures. But, as you already mentioned previously, the vapor pressure at the boiling point is equal to the ambient air pressure... so they have the same values.

 

[H.fulls]

so the vapour pressure at boiling point is the pressure I originally gave in my question?

 

[gneill]

so the vapour pressure at boiling point is the pressure I originally gave in my question?

Yes, that's one of them (the pressure at the high altitude location for which you want to find the corresponding boiling temperature). What's the other pressure/temperature pair?

 

[H.fulls]

Ahh okay.. is the other pair the boiling temperature and pressure for normal conditions? i.e 100 degrees and 101 KPa ?

 

[gneill]

Ahh okay.. is the other pair the boiling temperature and pressure for normal conditions? i.e 100 degrees and 101 KPa ?

Yes, of course.

 

[H.fulls]

Ahhhh okay! I finally understand! Thanks so much for your help :) !