Question

An absorption refrigeration system that receives heat from a source at \(95^{\circ} \mathrm{C}\) and maintains the refrigerated space at \(0^{\circ} \mathrm{C}\) is claimed to have a COP of \(3.1 .\) If the environmental temperature is \(19^{\circ} \mathrm{C}\), can this claim be valid? Justify your answer.

Short answer

Answer: Yes, the claim is valid, as the calculated Carnot COP (3.677) is higher than the given COP (3.1), indicating that the system operates within the maximum theoretical efficiency.

Step-by-step solution

Convert temperatures to Kelvin

Since all given temperatures are in degree Celsius, we need to convert them to Kelvin using the formula \(K = ^\circ C + 273.15\). Therefore, Heat source temperature (T₁) = \(95^\circ C + 273.15 = 368.15\,\text{K}\) Refrigerated space temperature (T₃) = \(0^\circ C + 273.15 = 273.15\,\text{K}\) Environmental temperature (T₂) = \(19^\circ C + 273.15 = 292.3\,\text{K}\)

Calculate Carnot COP

The COP of an ideal refrigeration system is governed by the highest theoretical efficiency possible, also known as Carnot COP. Using the refrigeration Carnot COP formula, we can calculate it as: Carnot COP = \(\frac{T_3}{T_1 - T_3}\) Plug in our values: Carnot COP = \(\frac{273.15}{368.15 - 273.15} \approx 3.677\)

Compare given COP with Carnot COP

Since the given COP is \(3.1\) and the calculated Carnot COP is approximately \(3.677\), the given claim of a COP of \(3.1\) is less than the maximum theoretical efficiency possible (Carnot COP), and therefore, the claim is valid.