A system is provided 50 J of heat and work don the system is 10 J. The change in internal energy during the process is :

Since the system is being worked on, the value of W will be positive since the system is receiving energy.

So, $\mathrm{W}=10\mathrm{J}.$ 

Also, heat provided is 50J which means q is also positive and $\mathrm{q}=50\mathrm{J}.$ 

So, using the first law of thermodynamics i.e., 

Change in internal energy = heat provided + work done

$\begin{array}{l}\Rightarrow \mathrm{\Delta }U=q+W\\ \Rightarrow \mathrm{\Delta }U=50J+10J=60J\end{array}$

$\mathrm{\Delta }U=q+w=50+10=60$