Question

1)A gas is compressed from a volume of 2m³ to a volume of 1m³ at a constant pressure of 100 N/m². Then it is heated at constant volume by supplying 150 J of energy. What happens to the internal energy of the gas?
2)Water of volume 2L in a closed container is heated with a coil of 1kW. While water is heated, the container loses energy at a rate of 160J/s. In how much time will the temperature of water rise from 27°C to 77°C? Specific heat of water is 4.2kJ/kg and that of the container is negligible.

Answer 1

1)  In the first step when the gas is compressed, perhaps no heat is supplied from outside the system (gas). ΔQ = 0.
      So ΔU = W = P ΔV = 100 * (-1) = -100 J
   In the 2nd step at constant volume, there is no work done. Hence heat supplied is equal to the increase in internal energy. 
      ΔU = 150 J.
   So finally the internal energy increases by 50 J from the beginning.

2)  V = 2 litres   density = 1 kg/ litre   mass = 2 kg
     Power = 1 kW   and   loss of energy = 160 W
    So net power = 840 W
    Time duration = energy/power = 2 kg * 4.2 * 1000 * (77-27) /840 sec
                           = 500 sec = 8 min 20 sec
    (we ignored the conductivity factor of water, ie., time taken for heat to 
      conduct from coil to the container through water)