A biology laboratory is maintained at a constant temperature...

A biology laboratory is maintained at a constant temperature of $7.00^{\circ} \mathrm{C}$ by an air conditioner, which is vented to the air outside. On a typical hot summer day the outside temperature is $27.0^{\circ} \mathrm{C}$ and the air conditioning unit emits energy to the outside at a rate of 10.0 $\mathrm{kW}$ . Model the unit as having a coefficient of performance equal to 40.0$\%$ of the coefficient of performance of an ideal Carnot device. (a) At what rate does the air conditioner remove energy from the laboratory? (b) Calculate the power required for the work input. (c) Find the change in entropy produced by the air conditioner in 1.00 h. (d) What If? The outside temperature increases to $32.0^{\circ} \mathrm{C} .$ Find the fractional change in the coefficient of performance of the air conditioner.

Key Concepts

Impact of Environmental Conditions on Device Performance

The performance of a refrigeration device is highly sensitive to the temperature difference between the cold space and the external environment. Changes in the external temperature affect the efficiency as measured by the COP; a higher external temperature increases the load on the device and typically results in a lower COP, thus mandating more work input for the same cooling effect.

Temperature Conversion in Thermodynamics

Accurate thermodynamic calculations require the conversion of temperature measurements from degrees Celsius to Kelvin, since absolute temperatures are fundamental to formulas involving thermal efficiency and COP. This conversion is critical to ensure that temperature differences are correctly represented in the thermodynamic equations.

Heat Transfer and Work Input

In refrigeration processes, heat transfer refers to the removal of unwanted thermal energy from a space, while work input is the external energy required to drive the process. The relationship between these is encapsulated in the COP, making it essential for determining both the rate of heat extraction and the effective power consumption of the cooling device.

Entropy Production in Irreversible Processes

Entropy production is a measure of the irreversibility of real processes. In any real refrigeration cycle, entropy increases due to irreversible phenomena, such as friction and non-ideal heat transfers, contrasting with the ideal (Carnot) cycle which assumes zero net change in entropy. This concept is vital for understanding the efficiency losses and thermodynamic cost of operating real-world systems.

Carnot Cycle

The Carnot cycle represents an ideal reversible process that establishes the maximum theoretical efficiency any heat engine or refrigerator can achieve between two temperature reservoirs. For refrigerators or air conditioners, the ideal COP is determined by the temperatures of the cold and hot reservoirs, emphasizing that no real device can exceed the performance set by the Carnot limit.

Energy Conservation in Thermodynamic Systems

Energy conservation, expressed through the first law of thermodynamics, states that energy cannot be created or destroyed but only transformed. In the context of a refrigeration cycle, the heat removed from the cold reservoir plus the work input must equal the heat rejected to the hot reservoir, ensuring a balance of energy flows within the system.

Coefficient of Performance (COP)

The coefficient of performance is a measure of the efficiency of a refrigeration device, defined as the ratio of the useful energy removed from the cold reservoir to the work input needed to accomplish that removal. It quantifies how much cooling effect is achieved per unit of work and is a critical parameter in evaluating the effectiveness of devices like refrigerators and air conditioners.

Transcript

00:01 Hello everyone in this problem it is given to maintain biology laboratory temperature constant at 7 degrees celsius.

00:12 It is to be converted in kelvin that is 280 kelvin and air condition is to be used.

00:26 Air condition is vented outside and hot summer day the outside temperature is 27 degrees celsius that is 300 kelvin if air condition ejecting the heat outside at the rate of that is qh by t is given 10 kilowatt and cofficent of performance actual coefficient of performance is 40 % of actual performance is 40 % of cofficent performance of card not cycle i'm correct this symbol coefficient of performance actually it is 40 % of that is 0 .4 times of coefficient of performance for carnot cycle this is given to us we have to find at what rate does the air condition remove energy from the laboratory so we have to calculate you see upon t in the part a in the part b power required we have to calculate to do the work input so input power we have to calculate and in c part find the change in entropy produced by the air condition change in entropy produced by the air condition in one hour and d if temperature outside becomes 32 degrees celsius then fractional change fractional change in coefficient of performance of the car not coefficient performance of air condition let us start solving with first part coefficient of performance of card not cycle is temperature of cold junction minus temperature of hot minus temperature of cold junction it is given 280 300 minus 280 so you will get 40 .0.

04:29 And actual performance of the air condition.

04:37 This is the actual performance.

04:39 It is 40%.

04:42 So 0 .4 into 40, that is 5 .6.

04:50 And it is defined as qc upon qh minus qc.

05:06 So from here you will get qc.

05:16 Qc be required to calculate from here.

05:27 On substituting the value, qc you will get 5.

05:34 5 .6 to be equal to qc upon t upon qh upon t minus qc upon t and this is given 10 kilowatt.

05:53 So you will get this value we have to calculate.

06:13 So on solving it, qc upon t you will get 8 .48 kilowatt.

06:23 So this is the amount of the heat.