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For a process where the internal energy increases by 41.4 J and 81.2 J of work is done by the system on its surroundings, what is the amount of heat associated with this process and its direction?

a) 122.6 J, released
b) 122.6 J, absorbed
c) 39.8 J, released
d) 39.8 J, absorbed

User Genene
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1 Answer

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Final answer:

The amount of heat associated with the process is 122.6 J, which is absorbed by the system. This is determined using the first law of thermodynamics (ΔU = Q - W), where the internal energy increase and the work done by the system on the surroundings are considered.

Step-by-step explanation:

To find out the amount of heat associated with the process where the internal energy increases by 41.4 J and 81.2 J of work is done by the system on its surroundings, we can use the first law of thermodynamics. This law is given by the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system on the surroundings. Plugging in our values, we get 41.4 J = Q - 81.2 J. Solving for Q gives us Q = 41.4 J + 81.2 J = 122.6 J. Thus, the heat associated with this process is 122.6 J, and since the internal energy of the system increased, this heat is absorbed by the system. Therefore, the correct answer is b) 122.6 J, absorbed.

User Molingran
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