Photo AI
Question 4
Figure 6 shows a low-voltage solid-state thermoelectric cooling element. The element is a square of side 40 mm and is 4 mm thick. Figure 7 shows how the element is ... show full transcript
Step 1
Answer
To determine the validity of the claim regarding the coefficient of performance (COP) of the thermoelectric refrigerator, we start by calculating the COP.
The COP can be calculated using the formula:
ext{COP} = rac{Q_C}{P}
where:
Substituting the values:
ext{COP} \ = 2.32$$ For an ideal refrigerator, the COP can be estimated using: $$ ext{COP}_{ ext{ideal}} = rac{T_C}{T_H - T_C}$$ Where $T_C$ and $T_H$ are the absolute temperatures of the cold and hot sides, respectively. Converting Celsius to Kelvin: - $T_C = 5 °C + 273.15 = 278.15 ext{ K}$ - $T_H = 35 °C + 273.15 = 308.15 ext{ K}$ Now, substituting into the ideal COP formula: $$ ext{COP}_{ ext{ideal}} = \frac{278.15}{308.15 - 278.15} \ = \frac{278.15}{30} \ ≈ 9.27$$ Comparing the declared COP value (2.32) to the ideal COP (9.27), it’s evident that the actual COP is significantly lower than the ideal value. Hence, the claim that the COP of a thermoelectric refrigerator is much less than the COP of an ideal refrigerator is valid.Step 2
Answer
One reason a small COP may be acceptable in this application is that thermoelectric coolers are compact and portable, making them suitable for specific uses such as medical applications where space is limited. They do not have moving parts, which results in high reliability and low maintenance. Additionally, the ability to maintain specific temperatures in isolated environments can be crucial for the preservation of sensitive medical supplies. Lastly, for applications requiring intermittent cooling, a lower COP could be manageable, provided that energy efficiency is balanced with the required performance for critical operations.
Report Improved Results
Recommend to friends
Students Supported
Questions answered