contestada

In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system. Find the amount of heat Q transferred in this process.
Express your answer numerically in kilojoules. Make your answer positive if the heat is transferred into the system; make it negative if the heat is transferred into the surroundings.

Respuesta :

Answer:

[tex]\Delta Q=-230kJ[/tex]

Explanation:

Using the first law of thermodynamics:

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

Where [tex]\Delta U[/tex] is the change in the internal energy of the system, in this case  [tex]\Delta U=250kJ[/tex], [tex]\Delta Q[/tex] is the heat tranferred, and [tex]W[/tex] is the work,  [tex]W=-480kJ[/tex] with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

[tex]\Delta Q=\Delta U +W[/tex]

And replacing the known values:

[tex]\Delta Q=250kJ +(-480kJ)[/tex]

[tex]\Delta Q=250kJ -480kJ[/tex]

[tex]\Delta Q=-230kJ[/tex]

The negative sign shows us that the heat is tranferred from the system into the surroundings.

Otras preguntas