Question

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occurs into the system, and 6.00 ✕ 106 J of heat transfer occurs to the environment?

Answer 1

The change in internal energy (ΔU) of the system is calculated using the first law of thermodynamics and is found to be -3.25 × 10^6 J, considering the work done and the net heat transfer.

Step-by-step explanation:

To calculate the change in internal energy of a system, you can use the first law of thermodynamics, which is expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat transfer into the system, and W is the work done by the system.

In this scenario, the system does 4.50 × 105 J of work (W = 4.50 × 105 J), there is a heat transfer of 3.20 × 106 J into the system (Qin = 3.20 × 106 J), and there is 6.00 × 106 J of heat transfer to the environment (Qout = 6.00 × 106 J).

The net heat transfer into the system is Q = Qin - Qout, which is 3.20 × 106 J - 6.00 × 106 J = -2.80 × 106 J. Now apply this to the first law of thermodynamics equation:

ΔU = Q - W = -2.80 × 106 J - 4.50 × 105 J = -2.80 × 106 J - 0.45 × 106 J = -3.25 × 106 J.

Therefore, the change in internal energy of the system is -3.25 × 106 J.

Answer 2

`-3.25* 10^6 J`

Step-by-step explanation:

We are given that

Work done by the system=
`4.5* 10^5` J

Heat transfer into the system=
`U_1=3.2* 10^6` J

Heat transfer to the environment=
`U_2=6* 10^6` J

We have to find the change in internal energy

By first law of thermodynamics

`\Delta Q=\Delta U+w`

`\Delta Q=U_1-U_2=3.2* 10^6-6* 10^6=-2.8* 10^6J`

Substitute the values then we get

`-2.8* 10^6=\Delta U+4.5* 10^5`

`\Delta U=-2.8* 10^6-4.5* 10^5=-28* 10^5-4.5* 10^5=-32.5* 10^5=-3.25* 10^6 J`

Hence, the change in internal energy =
`-3.25* 10^6 J`