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Heat and Thermodynamics by Anandamoy Manna

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Solved Problems

Q1. 1 gm of ice at normal pressure is changed in to 1 gm of water at 0°C. What is the change in the internal energy?

 

Ans. We have

 

dU = dQdW

 

Now,                                        dQ = 80 × 4.2 × 107 = 3.4 × 109 ergs

 

image

 

∴           dU = (3.4 × 103 + 0.092)×106 = 34×108 ergs

 

Q2. 1 cc of water at 100°C and normal pressure is changed into 1 gm of steam whose volume is 1649 cc at the same temperature and pressure. Find the change in internal energy.

 

Ans.dQ = 536 × 4.2 × 107= 2.25 × 1010 ergs

 

dW = 76 × 13.6 × 981 × 1649 = 1.67 × 109ergs

 

dW = dQdW = 2.25 × 1010 − 1.67 × 109 = 20.83 × 109ergs

 

Q3. A mass ...

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