A household refrigerator uses isobutane (C₄H₁₀) as the working fluid
PHYSICS · Multi-part · A Level · Physics · Further mechanics and thermal physics · Further mechanics and thermal physics
Problem
A household refrigerator uses isobutane (C₄H₁₀) as the working fluid. Inside the refrigerator coil, liquid isobutane at –10 °C vaporizes completely at a pressure where its boiling point is –10 °C. Outside the refrigerator, the vapor condenses at 30 °C and releases heat to the surroundings.
Data for isobutane:
• Molar enthalpy of vaporization at –10 °C = 22.5 kJ mol⁻¹
• Molar enthalpy of vaporization at 30 °C = 21.3 kJ mol⁻¹
• Molar mass = 58.1 g mol⁻¹
During steady operation, the refrigerator must remove 4.50 × 10² J of heat per second from the food compartment (assume contents remain at 0 °C).
For comparison, consider an experimental design that would instead use water vapor condensing at 30 °C to remove heat from the compartment at 0 °C.
Data for water at 30 °C: molar enthalpy of vaporization = 43.4 kJ mol⁻¹; molar mass = 18.0 g mol⁻¹
(a) Calculate the heat removed, in kJ, by the vaporization of 1.00 g of isobutane at –10 °C.
(b) Determine the mass flow rate, in g s⁻¹, of isobutane required to maintain removal of 4.50 × 10² J s⁻¹ from the food compartment.
(c) Calculate the mass flow rate, in g s⁻¹, of water that would be required if water vaporization/condensation at 30 °C were used instead to remove the same amount of heat.
(d) Evaluate the feasibility of using water rather than isobutane in this refrigeration cycle, citing at least two distinct thermodynamic or practical considerations revealed by your calculations.