Question

The temperature of a 12 -oz \((0.354-\mathrm{L})\) can of soft drink is reduced from 20 to \(5^{\circ} \mathrm{C}\) by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at \(20^{\circ} \mathrm{C}\) to the surroundings. There are no other heat transfers. Determine the minimum theoretical work input required by the cycle, in \(\mathrm{kJ}\), assuming the soft drink is an incompressible liquid with the properties of liquid water. Ignore the aluminum can.

Short answer

The minimum theoretical work input required is 436.26 kJ.

Step-by-step solution

- Determine the heat removed from the soft drink

Calculate the thermal energy removed from the soft drink. Use the formula for heat transfer: \[ Q = m \times c \times \Delta T \]where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Given: \( m = 0.354 \mathrm{kg} \) (since 1L of water is approximately 1kg), \( c = 4.18 \, \mathrm{kJ/kg.K} \) (specific heat capacity of water), and \( \Delta T = 20 ^{\circ} \mathrm{C} - 5 ^{\circ} \mathrm{C} = 15 ^{\circ} \mathrm{C} \).Thus, \[ Q = 0.354 \times 4.18 \times 15 = 22.2486 \, \mathrm{kJ} \]

- Determine the Carnot efficiency

The minimum theoretical work input is defined by the Carnot efficiency. The Carnot efficiency (\( \eta_{\text{Carnot}} \)) is given by: \[ \eta_{\text{Carnot}} = 1 - \frac{T_{L}}{T_{H}} \]Here, \( T_{L} = 5 ^{\circ} \mathrm{C} = 278.15 \, \mathrm{K} \) (Kelvin), and \( T_{H} = 20 ^{\circ} \mathrm{C} = 293.15 \, \mathrm{K} \). Thus, \[ \eta_{\text{Carnot}} = 1 - \frac{278.15}{293.15} = 0.051 \]

- Find the minimum theoretical work input

The work input (\( W_{\text{in}} \)) is related to the heat removed (\( Q \)) by: \[ W_{\text{in}} = \frac{Q}{\eta_{\text{Carnot}}} \]Substituting the values: \[ W_{\text{in}} = \frac{22.2486 \, \mathrm{kJ}}{0.051} = 436.26 \, \mathrm{kJ} \]

Key concepts

heat transfer

Heat transfer is the process of thermal energy moving from one object to another. In this problem, we decrease the temperature of a soft drink can from 20°C to 5°C through heat transfer. This involves calculating the quantity of heat removed. The formula used is:
\ Q = m \times c \times \Delta T .
\(m\) is the mass of the soft drink, \(c\) is its specific heat capacity, and \( \Delta T \) is the temperature change.
In simple terms, this equation tells us the amount of energy, in kilojoules (kJ), that needs to be transferred to cool the drink. In this case, the values are:

• \(m = 0.354 \) kg (since the soft drink behaves like water)
• \(c = 4.18 \mathrm{kJ/kg.K} \) (specific heat capacity of water)
• \( \Delta T = 15 ^{\circ} \mathrm{C} \).

\ Q \= 22.2486 \mathrm{kJ}.
This result indicates the thermal energy removed from the soft drink to bring its temperature down.

Carnot efficiency

Carnot efficiency describes the maximum theoretically possible efficiency of a heat engine operating between two given temperatures. For our refrigeration cycle, it illustrates how well we can convert heat energy into work.
The formula is:
\[ \eta_{\text{Carnot}} = 1 - \frac{T_{L}}{T_{H}} \] \[ T_{L} \text{ is the low temperature in Kelvin (278.15K)} \] \[ T_{H} \text{ is the high temperature in Kelvin (293.15K)} \] Substituting the temperatures into the Carnot efficiency equation gives:

\[ \eta_{\text{Carnot}} = 1 - \frac{278.15}{293.15} = 0.051 \].
This means that in the best-case scenario, our refrigeration cycle can only achieve an efficiency of around 5.1%. This inefficiency makes us understand the need for significant input energy to achieve cooling.

refrigeration cycle

A refrigeration cycle is a process that moves heat from a cooler location to a warmer one. It involves work input to carry energy from a low-temperature reservoir to a high-temperature one. In this exercise, the cycle cools a soft drink can from 20°C to 5°C.

In practical terms, the refrigeration cycle extracts 22.2486 kJ of heat from the soft drink. Recalling our Carnot efficiency, we calculate the minimum work input required using:
\ \[ \ W_{\text{in}} = \frac{Q}{\eta_{\text{Carnot}}} \ \]

• Q = 22.2486 kJ
• \eta_{\text{Carnot}} \approx 0.051.
\ Hence, the required work input is:
\ \ W_{\text{in}} = 436.26 \mathrm{kJ}.
This large value shows the inherent inefficiency in real-world refrigeration cycles.

specific heat capacity

Specific heat capacity \((c)\) is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius. For water, which the soft drink approximates, it is 4.18 kJ/kg.K.

In the exercise, we used this property to determine the thermal energy removed from the soft drink as its temperature was reduced by 15°C.
The formula used:
\ Q = m \times c \times \Delta T \

• m = 0.354 kg
• c = 4.18 kJ/kg.K
• \Delta T = 15^{\circ}C

Hence:
\ \ Q = 22.2486 \mathrm{kJ}.
Specific heat capacity plays a pivotal role in determining how much energy is needed for such temperature changes.