Question

Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsysΔSsysDeltaS_sys of the system? Assume that the temperatures of the objects do not change appreciably in the process

Answer 1

Explanation:

The given data is as follows.

       $T_{1}$ = 400 K,     $T_{2}$ = 500 K,  

$\Delta Q$ = 25 kJ = 25000 J   (as 1 kJ = 1000 J)

Now, we will calculate the change in entropy as follows.

            $\Delta S = \frac{\Delta Q}{\Delta T}$

Putting the given values into the above formula as follows.

           $\Delta S = \frac{\Delta Q}{\Delta T}$

                      = $\frac{25000 J}{(500 K - 400 K)}$      

                      = 250 J/K

Hence, we can conclude that change in entropy is 250 J/K.