# Answer to Question #16678 in Mechanics | Relativity for TomsMalaya

Question #16678

A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4 kg lid. If the local atmospheric pressure is 101 kPa, determine the temperature at which the water will start boiling when it is heated.

Expert's answer

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.

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.

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.

F(p,in) = F(p, out) + F(wt) (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 ymmetric 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 (here used the symbol A(proj) ).

F(p, out) = P(atm)*A(proj) (2)

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

F(p, in) = P(in)*A(proj) (3)

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

F(wt) = m(lid)*g/g(c)& (4)

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

P(in)*A(proj) = P(atm)*A(proj) + m(lid)*g/g(c)& (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).

P(in) = P(atm) + m(lid)/A(proj) * g/g(c)& (6)

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

A(proj) = π/4*D² = 0.031416 m²& (7)

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

Finally, we can go to the Saturation Pressure Table in the Steam Tables to determine the saturation pressure at Pin.& 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.

Interpolation yields : T(boil) = 100.2127 °C

Answer: T(boil) = 100.2 °C

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.

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.

F(p,in) = F(p, out) + F(wt) (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 ymmetric 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 (here used the symbol A(proj) ).

F(p, out) = P(atm)*A(proj) (2)

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

F(p, in) = P(in)*A(proj) (3)

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

F(wt) = m(lid)*g/g(c)& (4)

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

P(in)*A(proj) = P(atm)*A(proj) + m(lid)*g/g(c)& (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).

P(in) = P(atm) + m(lid)/A(proj) * g/g(c)& (6)

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

A(proj) = π/4*D² = 0.031416 m²& (7)

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

Finally, we can go to the Saturation Pressure Table in the Steam Tables to determine the saturation pressure at Pin.& 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.

Interpolation yields : T(boil) = 100.2127 °C

Answer: T(boil) = 100.2 °C

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