The heat supplied ( Q ) to a thermodynamic system and the work done...

The rate of heat supplied is the slope of the $Q$ versus $t$ graph. Since it is a straight line passing through the origin $(0, 0)$ and the point ($4\text{ s}, 240\text{ J}$):
$\dfrac{dQ}{dt} = \dfrac{240\text{ J} - 0}{4\text{ s} - 0} = 60\text{ J/s} = 60\text{ W}$

Similarly, the rate of work done is the slope of the $W$ versus $t$ graph, passing through the origin and ($4\text{ s}, 80\text{ J}$):
$\dfrac{dW}{dt} = \dfrac{80\text{ J} - 0}{4\text{ s} - 0} = 20\text{ J/s} = 20\text{ W}$

According to the first law of thermodynamics in rate form:
$\dfrac{dQ}{dt} = \dfrac{dU}{dt} + \dfrac{dW}{dt}$

Substituting the calculated rates:
$60 = \dfrac{dU}{dt} + 20$

$\dfrac{dU}{dt} = 60 - 20 = 40\text{ W}$

The rate of change of internal energy is $40\text{ W}$.

Answer: $40\text{ W}$