COP = Qc/W. So express W in terms of COP and Qc and solve.Runaway said:Homework Statement
433J of heat is extracted from a massive object at 0[tex]\circ[/tex]C while rejecting heat to a hot reservoir at 19[tex]\circ[/tex]C.
What minimum amount of work will accomplish this? Answer in units of J.
Homework Equations
COP= Qc/(Qh-Qc) = Tc/(Th-Tc)
The Attempt at a Solution
433J * 273.15K/(292.15K-273.15K)=6224.944 joules
Yes. But you should explain your reasoning.Runaway said:W= 433j/(273.15k/(292.15k-273.15k)) = 30.119j?
A Carnot refrigerator is a theoretical device that uses the Carnot cycle to cool an object. It consists of two heat reservoirs and a working substance, typically a gas. The Carnot cycle is a thermodynamic cycle that involves isothermal and adiabatic processes to transfer heat from a low-temperature reservoir to a high-temperature reservoir.
The minimum amount of energy needed to cool an object with a Carnot refrigerator is equal to the heat energy that needs to be removed from the object. This is known as the Carnot efficiency and is given by the equation: η = 1 - T_c/T_h, where T_c is the temperature of the cold reservoir and T_h is the temperature of the hot reservoir.
The minimum amount of energy needed to cool an object with a Carnot refrigerator is calculated using the Carnot efficiency equation: η = 1 - T_c/T_h. The temperatures of the cold and hot reservoirs are measured and substituted into the equation to determine the efficiency. The minimum amount of energy is then found by multiplying the efficiency by the heat energy that needs to be removed from the object.
No, the minimum amount of energy needed to cool an object cannot be achieved in practice due to energy losses and inefficiencies in real-world refrigeration systems. The Carnot efficiency is a theoretical limit and is not achievable in real-world systems. However, the closer a refrigerator's efficiency is to the Carnot efficiency, the more efficient it is.
The amount of energy needed to cool an object increases with the temperature difference between the hot and cold reservoirs. This is because a larger temperature difference results in a lower Carnot efficiency, meaning more energy is needed to achieve the same cooling effect. In other words, the greater the temperature difference, the less efficient the Carnot refrigerator becomes.