Question

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 starts boiling when it is heated

Answer 1

Answer:

$T=100.2^0C$

Explanation:

The boiling point of the water depends only on the absolute pressure p. the atmospheric temperature is given by P=101000 pa and the pressure from the lid we calculate by dividing its weight with the surface area of lid. the weight is product of  mass and the gravitational acceleration. surface area is calculated by given diameter.

$D=20cm=20cm.\frac{1m}{100cm} =0.2m\\A=D^2\pi /4\\p_{l} =W/A=\frac{m.g}{D^2\pi/4 }$

The absolute pressure p is sum of atmospheric pressure and  the pressure from the weight of lid.

$p=p_{a} +p_{l} \\\\p=p_{a} +\frac{m.g}{D^2\pi /4} \\\\p=101000Pa+\frac{4kg.9.81m/s^2}{(0.2m^2)^2.\pi /4} \\p=102249Pa$

At this pressure the boiling temperature of water is 100.2.

Answer 2

The atmospheric pressure table, the approximate temperature at an absolute pressure of 102.3kpa is 100 degrees

The given parameters are:

• $d = 20cm$ --- the diameter
• $m = 4kg$ --- the mass of the lid
• $p = 101kPa$ ---the local atmospheric pressure

Start by calculating the inner area (A) of the cooking pan

$A = \frac{\pi d^2}{4}$

So, we have:

$A = \frac{3.14 \times (20cm)^2}{4}$

Express 20cm as meters

$A = \frac{3.14 \times (0.20m)^2}{4}$

So, we have:

$A = 0.031m^2$

Next, we calculate the absolute pressure at this point

$P_a = P + \frac{W}{A}$

Where

$W= mg$ --- the weight of the lid

This gives

$Pa = P + \frac{mg}{A}$

$Pa = 101kpa + \frac{4 \times 9.8}{0.031}$

$Pa = 101kpa + 1264.52pa$

Express pa as kpa

$Pa = 101kpa + 1.26452kpa$

$Pa = 102.26452kpa$

Approximate

$Pa = 102.3kpa$

Using the atmospheric pressure table, the approximate temperature at an absolute pressure of 102.3kpa is 100 degrees

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