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kherson [118]
3 years ago
8

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

Physics
1 answer:
notsponge [240]3 years ago
6 0
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U 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
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
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