water in a 5-cm deep pan is observed to boil at 98 °c. at what temperature will the water in a 40-cm deep pan boil? assume both pans are full of water and take density of water to be 1000 kg/m3.
Answers
Given
Boiling temperature of water in 5 cm deep pan, T1 = 98 oC
In the first case, as the water starts boiling at 98 oC, the pressure at the bottom of the 5 cm deep pan will be the saturation pressure at that temperature (i.e., 98 oC) which can be obtained from the saturated water-temperature table.
From table we have
At T = 95 oC
Psat = 84.609 kPa
At T = 100 oC
Psat = 101.42 kPa
The saturation pressure at the given temperature 98 oC can be obtained by interpolating between the above two values as
Psat -84.609/101.42 - 84.609 = 98-95/100-95
Therefore, P1 = Psat @ 98 oC = 94.695 kPa
Now, consider 40 cm deep pan.
The pressure at the bottom of 40 cm deep pan = pressure at the bottom of 5cm deep pan + pressure due to the extra height (35 cm) of water
Therefore, P2 = P1 + pgh
Where p = density of water = 1000 kg/m3
h = difference in heights = 35 cm = 0.35 m
Thus,
P2 = 94.695 + 1000x9.81x0.35/1000 kPa
P2 = 98.13 kPa
Thus the boiling temperature (T2) of water in 40 cm deep pan will be the saturation temperature at the pressure 98.13 kPa which can be obtained from the saturation water-pressure table.
From table A-5, we have
At P = 75 kPa
Tsat = 91.76 oC
At P = 100 kPa
Tsat = 99.61 oC
The saturation temperature at the given pressure 104.85 kPa can be obtained by interpolating between the above two values as
T sat - 91.76/99.61 -91.7/99.61 - 91.76 = 98.13 - 75/100 - 75
Therefore T2 = Tsat @ 104.85k Pa
T2 = 99.022 degrees centigrade is the answer.