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A vessel whose walls are thermally insulated contains 2.40kg of water and 0.450kg of ice, all at 0.0C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to28.0C ? You can ignore the heat transferred to the container.

Short Answer

Expert verified

The mass of steam is 190g.

Given: Mass of ice m1=0.450kg, Mass of water is m2=2.40kginitial temperature and final temperature of system are

T1=0.0Cand Tf=28.0Crespectively.

Step by step solution

01

The conservation of heat energy

The heat gain by gain of ice and water is equals to the heat lost by steam.

Qice+Qwater=-Qsteam

Where, Qiceand Qwaterare heat gain by ice and water and Qsteamis heat lost by steam.

02

 The heat gain by ice, water and the heat lost by steam

The heat gain by ice is

225.6103J/kg

Qice=m1Lf+m1c1T1

Where, Lflatent heat of fusion of ice and c1is specific heat capacity of water having values 33.4103J/kgand 4190J/kg.Krespectively.

The heat gain by water is

Qwater=m2c1T2

Where, c1is specific heat capacity of water which is equals to 4190J/kg.K.

The heat lost by steam is

Qsteam=m3Lv+m3c1T3

Where, m3is the mass of steam and Lvis the latent heat of vaporization water which is equals to .

03

Calculation of mass of steam

Qice+Qwater=-Qsteamm1L1+m1c1T+m2c1T2=-m3Lv+m3c1T3m3=-m1Lf+m1c1T1+m2c1T2Lv+c1T3

Now, putting the values of constants in above equation

m3=-0.4533.4103+0.45419028-0+2.4419028-0225.6103+419028-100m3=0.190kgm3=190g

Thus, the mass of steam is 190g

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