## A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

Question

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## Answers ( 1 )

We can solve the problem by using the first law of thermodynamics, which states that:

[tex]Delta U = Q-W[/tex]

where

[tex]Delta U[/tex] is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is

[tex]Delta U= Q-W=+194 kJ – (-120 kJ)=+314 kJ[/tex]