heat in the amount of 5 KJ is added while its internal energy decreases by 10 KJ. how much energy is transferred as work? for a process causing the same change of state but for which the work is zero, how much heat is transferred?
To find the amount of energy transferred as work, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system:
ΔU = Q - W
Given that the heat added is 5 KJ and the change in internal energy is -10 KJ, we can substitute these values into the equation:
-10 KJ = 5 KJ - W
Rearranging the equation to solve for W, we find:
W = 5 KJ - (-10 KJ)
W = 5 KJ + 10 KJ
W = 15 KJ
Therefore, the energy transferred as work is 15 KJ.
For a process causing the same change of state but with zero work (W = 0), we can again use the first law of thermodynamics:
ΔU = Q - W
Since W is zero, the equation simplifies to:
ΔU = Q
Given that the change in internal energy (ΔU) is -10 KJ, we can substitute this value into the equation:
-10 KJ = Q
Therefore, in this case, the heat transferred (Q) is equal to -10 KJ.
To determine the amount of energy transferred as work, we need to understand the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat transfer (Q) into the system minus the work (W) done by the system:
ΔU = Q - W
Given that the internal energy decreases by 10 KJ (ΔU = -10 KJ) and 5 KJ of heat is added (Q = 5 KJ), we can rearrange the equation to solve for the work:
-10 KJ = 5 KJ - W
W = 5 KJ + 10 KJ = 15 KJ
Therefore, 15 KJ of energy is transferred as work.
For a process where the work is zero (W = 0), we can modify the first law of thermodynamics equation as:
ΔU = Q
Since the internal energy change is still -10 KJ (ΔU = -10 KJ), the amount of heat transferred (Q) will also be -10 KJ.
Therefore, in a process causing the same change of state but with no work being done, 10 KJ of heat is transferred.