Heat and Thermodynamics

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 = dQ − dW

 

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

 

 

∴           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 = dQ − dW = 2.25 × 10¹⁰ − 1.67 × 10⁹ = 20.83 × 10⁹ergs

 

Q3. A mass ...