In This post you will get answer of Q.
figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be:
Solution:
In cyclic process ABCA,
$Delta ce{U_{cyclic}} = 0$
$ce{Q_{cyclic} = W_{cyclic}}$
$ce{Q_{AB} = Q_{bc} + Q_{CA} =}$ closed loop area.
$400 + 100+ Q_{CA} = frac{1}{2}times (2times 10^{-3}) times 4times 10^4$
$ce{400 +100 Q_{AC = 40}}$
$ce{Q_{AC} = 460 J}$
Solution:
In cyclic process ABCA,
$Delta ce{U_{cyclic}} = 0$
$ce{Q_{cyclic} = W_{cyclic}}$
$ce{Q_{AB} = Q_{bc} + Q_{CA} =}$ closed loop area.
$400 + 100+ Q_{CA} = frac{1}{2}times (2times 10^{-3}) times 4times 10^4$
$ce{400 +100 Q_{AC = 40}}$
$ce{Q_{AC} = 460 J}$