A large
pot has a diameter of 14 cm.
It is filled with water
and covered with a heavy lid that weighs 3.7 kg. At what temperature does the water begin to boil if ambient
pressure is 101.325 kPa? 


Read : 
The key to
this problem is to recognize that the total pressure at the surface of the
liquid water must be greater than 101 kPa before the water can boil because of the weight of the
lid. This is true whether there is an
air space between the liquid water and the lid or not. As the temperature of the contents of the
pot rises, the pressure will increase.
When the 1st bubble of water vapor forms, it will displace some
air. The displaced air will escape by
lifting the lid. 














Given: 
D 
0.14 
m 


P_{atm} 
101.325 
kPa 


m_{lid} 
3.7 
kg 
















Find: 
T_{boil} 
??? 
^{o}C 
















Assumptions: 
None. 

















Equations
/ Data / Solve: 


















The liquid water will
boil when it reaches the temperature at which the vapor pressure of the water
is equal to the pressure required to lift the lid and let some air escape. 











Let's begin by
determining the pressure within the pot required to lift the lid. This can be accomplished by writing a force
balance on the lid. See the diagram. 












The lid will lift
slightly and let some air escape when the upward force exerted by the gas
inside the pot just balances the sum of the weight of the lid and the
downward force due to atmospheric pressure on the outside of the lid. 









Eqn 1 







The fact that the lid
is not flat on top does not affect the solution of this problem, as long as
the lid is axially symmetric about its centerline. 

















All of the horizontal
components of the forces acting on the lid cancel each other out (vector sum
is zero). The downward force is the same regardless of the shape of the top
of the lid. Remember that pressure
always acts in the direction perpendicular or normal to a surface. So as the lid surface curves, the downward
component of the pressure force decreases.
But the total surface area of the pot increases. These two factors are equal and
opposite. The result is that the force
exerted by the outside atmosphere on the pot lid is the same as if the lid
were flat. The area of an equivalent
flat surface is called the projected area (I use the symbol Aproj). 







Eqn 2 











Following the same
logic, the upward force exerted by the air in the pot on the lid can be
determined using : 


















Eqn 3 











The only term left is the
weight of the pot lid. This is an
application of Newton's 2nd Law. 



















Eqn 4 











Now, we can substitute
Eqns 2, 3 & 4 into Eqn 1 : 


Eqn 5 











The goal is to
determine the pressure inside the pot when the lid lifts and the water boils,
so let's solve Eqn 5 for the
unknown P_{in}. 







Eqn 6 











The only
unknown quantity on the righthand side of Eqn 6 is the projected area.
We can calculate its value using : 





Eqn 7 

A_{proj} 
0.015394 
m^{2} 











At last, we can plug
numbers into Eqn 6 and
evaluate the pressure in the pot when the water boils. 

Just be sure to use
the unit conversion : 












1 kPa 
1000 
N/m^{2} 

g 
9.8066 
m/s^{2} 







g_{c} 
1 
kgm/Ns^{2} 



P_{in} 
103.68 
kPa 











Finally, we can go to
the Saturation Pressure Table
in the Steam Tables to
determine the saturation pressure at P_{in}. This is the temperature at which the water
in the pot will boil. 











Because 102.25 kPa is not an entry in the
Saturation Pressure Table, an interpolation is required. 











T_{sat} (^{o}C) 
P_{sat} (kPa) 








100.00 
101.42 








T_{boil} 
103.68 








105.00 
120.90 

Interpolation yields : 

T_{boil} 
100.5810 
^{o}C 










Verify: 
None. 


















Answers : 
T_{boil} 
100.6 
^{o}C 















