Question

What is the change in internal energy of a system which does 4.50×10⁵ J of work while 3.00×10⁶ J of heat transfer occurs into the system, and 8.00×10⁶ J of heat transfer occurs to the environment?

a) −4.50×10⁶ J
b) −2.50×10⁶ J
c) −1.50×10⁶ J
d) −7.50×10⁶ J

Answer 1

The change in internal energy of the system is 2.55×10⁶ J.

Step-by-step explanation:

The change in internal energy of a system can be calculated using the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat transfer (ΔQ) minus the work done on the system (ΔW). In this case, the heat transfer into the system is 3.00×10^6 J and the work done on the system is 4.50×10^5 J. The heat transfer to the environment is 8.00×10^6 J. Therefore, the change in internal energy is:

ΔU = ΔQ - ΔW = (3.00×10^6 J) - (4.50×10^5 J) = 2.55×10^6 J

The change in internal energy (ΔU) of a system can be calculated using the first law of thermodynamics, which is ΔU = Q - W, where Q is the heat added to the system, and W is the work done by the system.

In this case, the heat added to the system is 3.00×106 J, and the heat transferred to the environment is 8.00×106 J, so the net heat Q is -5.00×106 J (since 3.00×106 J - 8.00×106 J = -5.00×106 J). The work done by the system is 4.50×105 J. Therefore, the change in internal energy is ΔU = -5.00×106 J - 4.50×105 J = -5.45×106 J.

So, the change in internal energy of the system is 2.55×10^6 J (option b).