A scientist needs a refrigeration machine to maintain temperature of -...
**Solution:**
To find the work performed on the system, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
In this case, the heat added to the system is -3000J (since heat is being withdrawn from the reservoir), and the work done by the system is what we need to find.
**Finding the Change in Internal Energy:**
The change in internal energy can be calculated using the equation:
ΔU = mcΔT
where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, since the substance is not specified, we can assume a unit mass. The specific heat capacity of water is approximately 4.18J/g°C, so we can use this value as an approximation.
ΔU = mcΔT = (1g)(4.18J/g°C)(27°C - (-13°C)) = 1g(4.18J/g°C)(40°C) = 167.2J
Therefore, the change in internal energy is 167.2J.
**Calculating the Work Done:**
Now we can use the first law of thermodynamics to find the work done by the system:
ΔU = Q - W
167.2J = -3000J - W
W = -3000J - 167.2J = -3167.2J
Since work is a transfer of energy, it is a scalar quantity and can be negative. In this case, the negative sign indicates that work is being done on the system.
Therefore, the work performed on the system during each cycle of its operation is -3167.2J (or approximately -3167J).
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