Question

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.

Answer 1

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case  $\Delta U=250kJ$, $\Delta Q$ is the heat tranferred, and $W$ is the work,  $W=-480kJ$ 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

$\Delta Q=\Delta U +W$

And replacing the known values:

$\Delta Q=250kJ +(-480kJ)$

$\Delta Q=250kJ -480kJ$

$\Delta Q=-230kJ$

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