Question

1 kg of water at 373 K is converted into steam at same temperature. Volume of 1 cm" of water becomes1671 cm on boiling. What is the change in the internal energy of the system if the latent heat of vaporisationof water is 5.4 x 105 cal kg-l?Solution​

Answer 1

The change in the internal energy of the system is 2098.74 J

Explanation:

By using the first law of thermodynamics.

ΔQ = ΔU + ΔW

So, change in internal energy will be

ΔU = ΔQ - ΔW  

Here,

ΔQ = mL =1g x 540Cal/g = 540 Cal

ΔU  = change in internal energy (to be calculated)

ΔW = change in work done

So,   ΔW = pΔV  

Where

ΔV = 1671cm3 - 1cm3 = 1670cm3 = 1670 x 10-6 m3

p = hρg = 0.76 x 13600 x 9.8 ~ 101293 N/m2

ΔW = 101293 x 1670x10-6 = 169.16J

ΔW = 169.16 / 4.2 = 40.3 Cal

Thus, we have

ΔU = 540 Cal - 40.3 Cal

ΔU = 499.7 Cal = 2098.74 J

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