00:01
Hello, so in this question we need to calculate the amount of energy required to vaporize one gallon of water at 20 degrees celsius in beginning.
00:11
So if we assign the point a, here a point a is assigned as 20 degrees celsius.
00:19
Now temperature is increased from a to b where temperature is boiling, that is 100 degrees celsius.
00:29
Now from point b to point c, temperature remains constant.
00:32
But there is enthalpy of vaporization that is q of vaporization okay whereas from a to b there will be q of fusion that is enthalpy of fusion so firstly if we convert one gallon of water into grams so we will multiply by 3 .785 liters per one gallon times thousand ml per one liter times one gram per one m l so this liter this liter this liter cancelled out gallons and gallons cancelled out and mls get cancelled out and it comes out to be 3 .75 times 10 to power 3 grams this much is needed to be vaporized okay so now we are going to use the equation for enthalpia of fusion that is equals to m c s delta t where mass is given as 3 .75 times standards to power 3 grams so and the cs is given as specific heated constant temperature is given as 4 .18 4 .18 jule per gram degree celsius times delta is from 100 minus 20 degrees celsius.
02:15
Also if we convert it into kilojoules then it will be easy to calculate.
02:19
So we will convert it into kilojoules.
02:23
So one kilojoule per 10 is to power 3 joules.
02:28
So after solving this, enthalpy of fusion comes out to be 1 .267 times 10 is to power 3 kilojoules.
02:39
This is the enthalpy of fusion...