(a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of 50.0°C and a cold reservoir temperature of -20.0°C?
(b) How much heat in kilocalories would it pump into the warm environment if 3.60 multiplied by 107 J of work (10.0 kW·h) is put into it?
(c) Assume the cost of this work input is 10¢/kW·h. Also assume that the cost of direct production of heat by burning natural gas is 88.0¢ per therm (a common unit of energy for natural gas), where a therm equals 1.055 multiplied by 108 J. Compare the cost of producing the same amount of heat by each method. (cost of heat pump / cost of natural gas)
a. Tc=253.15k, Th=323.15
COP=1/(1-253.15/323.15)=4.62
To calculate the coefficient of performance (COP) for a heat pump, you need to know the temperatures of the hot and cold reservoirs. The COP is defined as the ratio of the heat transferred to the hot reservoir to the work input:
COP = Qh / W
where Qh is the heat transferred to the hot reservoir and W is the work input.
(a) In this case, the hot reservoir temperature is 50.0°C, which is equivalent to 323.15 K, and the cold reservoir temperature is -20.0°C, which is equivalent to 253.15 K. Plugging these values into the COP formula, we get:
COP = 1 / (1 - Tc / Th)
COP = 1 / (1 - 253.15 / 323.15)
COP = 1 / (1 - 0.7834)
COP ≈ 4.62
Therefore, the best coefficient of performance for this heat pump is approximately 4.62.
(b) To calculate the amount of heat in kilocalories that the heat pump would pump into the warm environment when a certain amount of work is put into it, you need to know the work input and the relationship between work and heat transfer. The relationship is given by the equation:
Q = W / COP
where Q is the heat transferred and COP is the coefficient of performance.
In this case, the work input is given as 3.60 x 10^7 J (10.0 kW·h), and we previously found that the COP is approximately 4.62. Plugging these values into the equation, we can solve for Q:
Q = (3.60 x 10^7 J) / 4.62
To convert the result from joules to kilocalories, we need to divide by 4.184 J/cal:
Q ≈ (3.60 x 10^7 J) / (4.62 x 4.184 J/cal)
Q ≈ 173,749 cal
Therefore, the heat pump would pump approximately 173,749 kilocalories of heat into the warm environment when 3.60 x 10^7 J of work is put into it.
(c) To compare the cost of producing the same amount of heat by using the heat pump versus burning natural gas, we need to determine the cost for each method.
For the cost of the work input for the heat pump, it is given as 10¢/kW·h. Since the work input is 10.0 kW·h, the cost for the heat pump can be calculated as:
Cost of heat pump = (10¢/kW·h) x 10.0 kW·h
Cost of heat pump = 100¢
For the cost of direct production of heat by burning natural gas, it is given as 88.0¢ per therm, where 1 therm is equal to 1.055 x 10^8 J. To find the cost for the same amount of heat produced by natural gas, we need to convert the heat transferred from kilocalories to J and then calculate the number of therms:
Q = 173,749 cal x 4.184 J/cal
Q ≈ 726,332 J
Number of therms = Q / (1.055 x 10^8 J/therm)
Number of therms ≈ 726,332 J / (1.055 x 10^8 J/therm)
Number of therms ≈ 0.00688 therms
Cost of natural gas = (88.0¢/therm) x 0.00688 therms
Cost of natural gas ≈ 0.61¢
Therefore, the cost of producing the same amount of heat using the heat pump is approximately 100¢, while the cost of producing the same amount of heat by burning natural gas is approximately 0.61¢.