Problem 57 - University Physics with Modern Physics

Question

20.57. (a) How much work must a Carnot refrigerator do on a hot day to transfer 1000 \(\mathrm{J}\) of heat from its interior at \(10^{\circ} \mathrm{C}\) to the out- side air at \(35.0^{\circ} \mathrm{C}\) ? (b) How much work must the same refrigerator do to transfer the same amount of heat if the interior temperature is the same, but the outside air is at only \(15.0^{\circ} \mathrm{C} ?\) (c) Sketch \(p V_{-}\) diagrams for these two situations. Can you explain in physical terms why more work must be done when the temperature difference between the two isothermal stages is greater?

Step-by-Step Solution

Answer

(a) 88.29 J, (b) 17.66 J. More work is needed when the temperature difference is greater due to reduced efficiency.

Step 1: Convert Temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin. This is done by adding 273.15 to the Celsius temperature.For the refrigerator interior at \(10^{\circ} \mathrm{C}\):\[ T_c = 10 + 273.15 = 283.15 \; K \]For the outside air at \(35^{\circ} \mathrm{C}\):\[ T_h = 35 + 273.15 = 308.15 \; K \]For the outside air at \(15^{\circ} \mathrm{C}\):\[ T'_h = 15 + 273.15 = 288.15 \; K \]

Step 2: Calculate Coefficient of Performance (COP) for Each Scenario

The coefficient of performance (COP) for a Carnot refrigerator is given by the formula:\[ \text{COP} = \frac{T_c}{T_h - T_c} \]For the first scenario with outside temperature at \(35^{\circ} \mathrm{C}\):\[ \text{COP}_1 = \frac{283.15}{308.15 - 283.15} = \frac{283.15}{25} = 11.33 \]For the second scenario with outside temperature at \(15^{\circ} \mathrm{C}\):\[ \text{COP}_2 = \frac{283.15}{288.15 - 283.15} = \frac{283.15}{5} = 56.63 \]

Step 3: Calculate Work Done by Refrigerator for Each Scenario

The work done by the refrigerator is related to the heat transferred \(Q_c\) and the COP as:\[ W = \frac{Q_c}{\text{COP}} \]Given that \(Q_c = 1000 \; J\):For the first scenario:\[ W_1 = \frac{1000}{11.33} = 88.29 \; J \]For the second scenario:\[ W_2 = \frac{1000}{56.63} = 17.66 \; J \]

Step 4: Sketch \(pV\) Diagrams and Explain Work Difference

Draw two \(pV\) diagrams, one for each scenario. The diagrams should display isothermal and adiabatic processes, where the isothermal expansion and compression occur at different temperatures for the two scenarios.The greater work required when the temperature difference is larger (as in the first scenario) is due to a lower efficiency in transferring heat at higher temperature differences in a Carnot cycle. This requires more input work to achieve the same heat transfer at a higher temperature gradient.

Key Concepts

Coefficient of Performance (COP)

The Coefficient of Performance, or COP, is a critical concept in understanding how efficiently a refrigerator operates. The COP of a refrigerator is defined as the ratio of the heat removed from the cold reservoir (\( Q_c \)) to the work done by the refrigerator (\( W \)). It is given by the formula:

• \( \text{COP} = \frac{T_c}{T_h - T_c} \)

where:

• \( T_c \) is the absolute temperature of the cold reservoir (inside the refrigerator), and
• \( T_h \) is the absolute temperature of the hot reservoir (the outside air).

The COP determines how much work is needed to transfer a certain amount of heat. A higher COP means the system is more efficient, requiring less work to move a given amount of heat. This happens when the temperatures of the inside and outside are closer together, as seen when the outside air was at \( 15^{\circ} \mathrm{C} \) compared to \( 35^{\circ} \mathrm{C} \). It’s important because a higher COP implies energy savings, making it a critical factor in designing efficient refrigeration systems.

Work done by refrigerator

The work done by a refrigerator is the energy input needed to transfer heat from inside the unit to the outside environment. This work can be calculated using the relationship between the heat removed (\( Q_c \)) and the Coefficient of Performance (COP). The formula for the work done by the refrigerator is:

• \( W = \frac{Q_c}{\text{COP}} \)

This means that if the COP is low, the work \( W \) required will be higher. For instance, less work was needed to transfer 1000 J of heat when the outside temperature was lower (\( 15^{\circ} \mathrm{C} \)) because the COP was higher at that temperature (\( 56.63 \)). Conversely, when the outside temperature was higher (\( 35^{\circ} \mathrm{C} \)), the COP was lower, and more work (\( 88.29 \, J \)) was needed. The efficiency of a refrigerator in doing work is crucial for energy conservation, as it directly relates to the energy costs of running the appliance.

Temperature conversion to Kelvin

Understanding temperature conversion from Celsius to Kelvin is essential for calculating various thermodynamic properties. The Kelvin scale is used in science for absolute temperature measurements because it starts from absolute zero, the lowest possible temperature (-273.15 °C).To convert Celsius to Kelvin, you simply add 273.15:

• \( T(K) = T(°C) + 273.15 \)

In the given exercise, converting the refrigerator's internal temperature of \( 10^{\circ} \mathrm{C} \) and the external temperatures of \( 35^{\circ} \mathrm{C} \) and \( 15^{\circ} \mathrm{C} \) into Kelvin was the first step. This conversion ensures that all thermodynamic calculations, like those for COP and Work, use an absolute temperature scale, which is crucial for accuracy. Without converting to Kelvin, calculations based on temperature differences or ratios could lead to incorrect results, as the Celsius scale does not account for absolute zero.