00:01
So this question it is given that suppose there is a 1 .15 gram of a rubbing alcohol that is c3hi2 and it evaporates from a 65 gram of aluminium block.
00:14
Now if the aluminium block is initially at 25 degrees celsius, then what is the final temperature of the block after the rubbing alcohol is evaporated? and here it is also given that we have to assume the heat required for the vaporization of the alcohol comes only from this aluminum block.
00:36
So to solve this question first we will take the heat that is required to evaporate the alcohol and the heat that is being given by the aluminum block and then we'll equate the both the heats to find the value of the finite temperature.
00:53
First we'll calculate heat to first we'll calculate the heat to evaporate the alcohol that is c3 h .a2.
01:00
So heat is equal to the number of moles into delta is preparation.
01:05
So the number of moles can be calculated by given mass of molar mass the given mass is 1 .15 grams and the molar mass of the alcohol is 60 .1 gram per more and the delta is vaporization it is 45 .4 kilojou so when we calculate we will get 0 .8682 kilojouze that is 868 .72 joules.
01:25
Now this is the heat to require to evaporate the alcohol.
01:30
Now heat being taken from the aluminum block to evaporate this alcohol so it can be calculated by mc delta.
01:37
That is the mass of the aluminum block into the specific heat capacity of the aluminum into the change in temperature...