Suppose you have a pot filled with steam at 110 $\circ$ C and at a pressure of 1 atm.The pot has a 20 cm diameter and is 15 cm high.You place a heavy stone on the lid,so that the combined mass of the lid and stone is 45 kg.How hot do you need to heat the steam to lift the lid off the pot?

Question

Quizplus · Accepted Answer

To lift the lid off the pot, the pressure inside the pot needs to be greater than 1 atm, which means you need to increase the temperature of the steam. We can use the ideal gas law to solve for the temperature needed:
PV=nRT 
Since the pot is closed, we can assume the volume is constant:
V=π(r^2)h=(π(10cm)^2)(15cm)=(1500π)cm^3
We can also assume that the mass of the lid and stone is evenly distributed over the lid, so that the pressure applied to the lid is:
P=F/A
where F=mg is the weight of the lid and stone, and A is the area of the lid:
A=π(r^2)=π(10cm)^2=100π cm^2
Therefore, the pressure applied to the lid is:
P=(45kg)(9.8m/s^2)/((100π)cm^2)=1.44 kPa
Since the pressure inside the pot needs to be greater than atmospheric pressure (101.3 kPa), we have:
P>101.3 kPa
Substituting these values into the ideal gas law:
nRT/P=V
Solving for T:
T=P(V/nR)=(1.44 kPa)((1500π)cm^3/(1 mol)/(8.31 J/(molK)))=176°F=80°C
Converting to Fahrenheit:
(9/5)(80°C)+32=163°F

Suppose You Have a Pot Filled with Steam at 110 ∘\circ∘

Suppose You Have a Pot Filled with Steam at 110^$\circ$