Determine the heat absorbed by a system undergoing the cyclic...

For a cyclic process, the change in internal energy is zero ($\Delta U = 0$).

According to the first law of thermodynamics, the heat absorbed is equal to the work done:

$Q = W$

The magnitude of work done in a cyclic process is given by the area enclosed by the cycle on the indicator diagram.

From the given $V-p$ graph, the semi-axes of the closed curve are:

$r_p = \dfrac{300 - 100}{2} = 100 \text{ kPa} = 10^5 \text{ Pa}$

$r_V = \dfrac{300 - 100}{2} = 100 \text{ cc} = 100 \times 10^{-6} \text{ m}^3 = 10^{-4} \text{ m}^3$

The area of the cycle is:

$\text{Area} = \pi \times r_p \times r_V$

$\text{Area} = \pi \times (10^5 \text{ Pa}) \times (10^{-4} \text{ m}^3)$

$\text{Area} = 10\pi \text{ J}$

Using $\pi \approx 3.14$, we get:

$|W| = 10 \times 3.14 = 31.4 \text{ J}$

Therefore, the magnitude of heat absorbed is $31.4 \text{ J}$.