Suppose a system returns to its original overall, internal energy ((∆u) after the following changes. in step 1, 25 j of work is done on the system as it releases 37 j of energy as heat. if, in step 2, the system performs 12 j of work, how much heat is transferred?

Heat transfer in step 2 = 47.75 J

Explanation:

Internal energy = heat + work done

U = Q + W

In a cyclic process the total internal energy change of the system = 0.

In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.

We have to find the heat exchange in step 2.

In step 1,

W = 1.25 J          Q = -37 J

= -37 + 1.25 = -35.75 J

In step 2, the internal energy change will be negative of that in step 1.

U = 35.75 J

W = -12 J

U = Q + W

35.75 = Q -12

Q = 47.75 J

Heat transfer in step 2 = 47.75 J

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-