00:01
In this question, it is told that on a winter day when temperature drops to minus 10 degrees celsius, ice is formed on the surface of a lake.
00:08
So first of all, i am writing the given data for this question.
00:12
We have given the thermal conductivity of ice that is 1 .7 watt per meter degrees celsius.
00:18
We have given the density of water that is 10 to the power 3 kilograms per meter cube.
00:23
And latent heat of fusion of ice that is alice is given to us 3 .36 into 10 to the power 5.
00:30
Jule per kg and length of ice form that is small l is 10 semi or i can say 10 into 10 to the power minus 2 meter now we in the first part of the question we have to find the rate of increase of thickness of the ice so we know that the rate of flow of heat is given by that is dlq by dlt is equals to this is k a into t1 minus t2 upon l.
01:13
Now dlq is equal to this is mass into latent heat or fusion of ice upon dlt is equal to this is ka t1 minus t2 upon small l.
01:28
Now from here we can say that l upon delta t is equal to this is ka t1 minus t2 upon mass can be written as row of water into area into small l.
01:43
This is mass into this is latent heat of fusion of ice now put all the data in this equation so rate of increase of thickness that is l upon delta t is given by this is 1 .7 into 0 minus this is minus 10 upon this is 10 to power 3 into this area and this area will cancel out this is 10 into 10 to the power minus 2 into 3 .36 into 10 to the power 5 after solver this we get that l upon delta t is equal to this is 5 into 10 to the power minus 7 meter per second so this is the rate of increase of forming of ice and this is our answer of first part now in the second part we have to find the total time taken to form a 10 semi -thick ice so let to form a thin layer of thickness dt to time is required.
03:04
So we can say that mass of thin layer is that is dm is equal to row of water into area into dx...