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

The change in internal energy for the overall process is 0 J.

Step-by-step explanation:

The change in internal energy of a system can be calculated using the equation:

ΔU = Q - W

Where ΔU is the change in internal energy, Q is the heat transferred to or from the system, and W is the work done on or by the system.

In this case, for the overall process, we have:

ΔU = Q1 + Q2 - W1 - W2

Substituting the given values:

ΔU = 222 J + 123 J - 150 J - 195 J

ΔU = 0 J

Therefore, the change in internal energy for the overall process is 0 J. This means that the heat transferred to the system is equal to the work done on the system.

Answer 2

Answer:0 J

Step-by-step 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