Question

A rigid container has water of 2 kg at 120 C with a quality of 25%. The container is heated to the temperature of 140 Celsius, calculate the new quality?

Answer 1

Given the mass of water = 2 kg = 2000 g

Also the acutal quality = 25 %

Therefore the actual mass of water , m = 2000 x 25%

                                                        = (2000x25) / 100

                                                        = 500 g

We know that PV = nRT

                    PV = (m/M) RT

Where

T = Temperature ;P = pressure ; n = No . of moles= mass(m)/molar mass(M) ;R = gas constant ;

V= Volume of the gas

As P , M & V are the same

So mT = constant

So mT = m'T'                           OR   m / m' = T' / T

Where

m = actual mass = 500 g

m' = new mass at T' = ?

T = initial temperature = 120 ^oC = 120+273 = 393 K

T' = final temperature = 140 ^oC = 140+273 = 413 K

Plug the values we get

m' = (mT )/ T'

    = (500x393) / 413

    = 475.8 g

Therefore the quality is (475.8 x 100) / 2000

                                  = 23.8 %