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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.

Respuesta :

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is [tex]\Delta U_1=222 J[/tex]

Work done on the system [tex]W_1=-150 J[/tex]

For second step

change in internal Energy of the system is [tex]\Delta U_2=123 J[/tex]

Work done on the system [tex]W_2=-195 J[/tex]

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

and from first law of thermodynamics

[tex]Q=\Delta U+W[/tex]

for first step

[tex]Q_1=222-150=72 J[/tex]

[tex]Q_2=123-195=-72 J[/tex]

overall heat added[tex]=Q_1+Q_2[/tex]

[tex]Q_{net}=72-72 =0[/tex]

For overall Process Heat added is 0 J

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