Question

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of work is done on the system. In the second step, the internal energy of the system increases by 123 J when 195 J of work is done on the system. For the overall process, find the heat.

Answer 1

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added$=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J