00:01
Hello students, hope you are doing great.
00:03
In this question we are given that 537 g of water in the liquid state is converted into water vapor.
00:10
So the temperature is raised from 0 degree celsius to 187 degree celsius.
00:15
So here it should be noted that at 0 degree celsius water is present in the solid state.
00:20
From solid state it is converted into liquid state and from liquid it is converted into the vapor phase.
00:26
So we have to find out the total heat energy required.
00:32
So from converting the water, from raising the temperature from 0 degree celsius to 100 degree celsius will require q1 and converting the water vapor into stream will require energy q2 and from raising the temperature from 100 degree celsius to 187 degree celsius will require q3 and total heat required is equals to q1 plus q2 plus q3.
01:03
So let us find out the values of q1, q2 and q3 all together.
01:08
And here the specific heat of water is given as 4 .184 joule per gram, latent heat of vaporization is equals to 2259 joule per gram and specific heat of vapor is equals to 2 .02 joule per gram.
01:31
So now we will first find out q1.
01:35
So q1 is equals to mass of water into specific heat of water into temperature change.
01:45
Mass of water is 537 g, specific heat is 4 .184 joule per gram degree celsius and change in temperature is from 0 degree celsius to 100 degree celsius.
01:57
So delta t is 100 degree celsius.
01:59
Calculating this we have 224 .68 kilo joule.
02:03
Now we will find out q2...
