40. A path 1-2-3 is given. A system absorbs 100 kj as heat and does 60kj of work while along the path 1-4-3, it does 20kj of work. The heat absorbed during the cycle 1-4-3 is-
For 1-2-3 cycle.
𝘘1-2-3 = 100 kj
W1-2-3 = 60 kj
We know,
Heat = Internal energy + Work
𝘘1-2-3 = 𝘘1-2-3 + W1-2-3
𝘘1-2-3 = 𝘘1-2-3 - W1-2-3
𝘘1-2-3 = (100-60) kj = 40 kj
Again, For 1-4-3 cycle,
W1-4-3 = 20kj
𝘘1-2-3 = 𝘘1-4-3 =40kj
Again, we know,
𝘘1-4-3 = 𝘘1-4-3 + W1-4-3
= 40 kj+20 kj = 60 kj
For 1-2-3 cycle.
𝘘1-2-3 = 100 kj
W1-2-3 = 60 kj
We know,
Heat = Internal energy + Work
𝘘1-2-3 = 𝘘1-2-3 + W1-2-3
𝘘1-2-3 = 𝘘1-2-3 - W1-2-3
𝘘1-2-3 = (100-60) kj = 40 kj
Again, For 1-4-3 cycle,
W1-4-3 = 20kj
𝘘1-2-3 = 𝘘1-4-3 =40kj
Again, we know,
𝘘1-4-3 = 𝘘1-4-3 + W1-4-3
= 40 kj+20 kj = 60 kj