Refrigeration units can be rated in “tons.” A 1-ton air conditioning system can remove sufficient energy to freeze 1 ton (2000 pounds = 909 kg) of 0°C water into 0°C ice in one 24-h day. Assume the hot part of a day averages 35°C and the interior of a house is maintained at 22°C by the continuous operation of a 6-ton air conditioning system for 6 hours a day. How much does this cooling cost the homeowner per day, and per month?Assume the work done by the refrigeration unit is powered by electricity that costs $0.13 per kWh and that the unit’s coefficient of performance is only 18% of an ideal refrigerator. 1 kWh = 3.60 x 10⁶ J .

Verified step by step guidance

1. Calculate the energy required to freeze 1 ton of water into ice. Use the latent heat of fusion for water, which is approximately 334 kJ/kg. Multiply this by the mass of the water (909 kg) to find the total energy required in kJ.
2. Convert the total energy required from kJ to kWh, knowing that 1 kWh = 3.60 x 10^6 J. This will give you the energy in kWh needed to freeze 1 ton of water.
3. Determine the actual energy used by the air conditioning system by considering its coefficient of performance (COP). The ideal COP for a refrigeration cycle can be approximated using the temperatures of the hot reservoir (35°C) and the cold reservoir (22°C). Calculate the ideal COP using the formula: COP_ideal = T_cold / (T_hot - T_cold), where temperatures are in Kelvin.
4. Adjust the ideal COP by the efficiency of the actual unit, which is 18% of the ideal. Multiply the ideal COP by 0.18 to find the actual COP of the unit.
5. Calculate the total energy consumed by the air conditioning system in kWh, considering it operates for 6 hours a day. Use the actual COP to find the energy input required by the system. Then, multiply the energy per day by the cost of electricity per kWh to find the daily cost. Multiply this by 30 to find the monthly cost.