Respuesta :
In this case we apply the concepts in thermodynamics. When it comes to intensive properties like internal energy (U), heat transfer (Q) and work (W), there are already derived relationships for this which were based on differential equations. The equation is
ΔU = Q + W
But we should remember to incorporate their signs because these parameters are dependent on these. According to Van Ness, Smith and Abbott, the sign convention for decreasing is negative. For work, the sign is positive if it is work ON the system, but negative if it is work done BY the system. So, in this case,
ΔU = -150 J
W = - 30 J
Then, we can determine Q.
-150 J = Q - 30 J
Q = - 120 J
This means that 120 J of heat was released by the system to the surroundings.
ΔU = Q + W
But we should remember to incorporate their signs because these parameters are dependent on these. According to Van Ness, Smith and Abbott, the sign convention for decreasing is negative. For work, the sign is positive if it is work ON the system, but negative if it is work done BY the system. So, in this case,
ΔU = -150 J
W = - 30 J
Then, we can determine Q.
-150 J = Q - 30 J
Q = - 120 J
This means that 120 J of heat was released by the system to the surroundings.
Hence the heat transfer that occurs from a system will be -120 J.
What is heat transfer?
Heat transfer refers to the movement of heat. The flow of heat across a system's boundary is due to a temperature differential between the system and its surroundings.
The given data in the problem is;
ΔU is the internal energy -150 J
W is the work fon = - 30 J
Q is the heat transfer that occurs from a system
The internal energy of the system is found as;
[tex]\rm \triangle U = Q + W \\\\-150 =Q-30 \\\\ Q=-120 \ J[/tex]
The heat transfer that occurs from a system will be -120 J.
To learn more about the heat transfer refer to the link;
https://brainly.com/question/13433948
