Question

Work done in converting one gram of ice at -10°c into steam at 100°c is

Answer 1

Answer:

Explanation:

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Answer 2

Work done in converting  1g of ice to stream is 3.039 joules.

We know that according to thermodynamics Q=mass*specific heat* temperature

So for ice Q = 1*2.09*10 = 20.9 joules.......(1)

For water Q = 1*4.18*100 = 418.6 joules.......(2)

So ice is melted so it has some heat hidden by which it gets converted which is known as heat of fusion =333 J/g

So we have Q=mass * heat of fusion

Q = 1*333 = 333 J........(3)

For water is boiled and converted to stream so some heat is stored in stream at 100 degree celcius which is known as heat of vaporization = 2260 J/g

So Q= 1*2260 = 2260 J........(4)

So from equations 1,2, 3, 4 we get total work done in converting 1g of ice at -10 to stream att 100 degree celcius as (1) +(2) +(3) +(4)

So = (20.9 + 4187.6 + 333 +2260) Joules = 3.0325 Joules

So work done is 3.0325 Joules.