(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) of a heat pump, you need to use the formula:
COP = Th / (Th - Tc),
where Th is the temperature of the hot reservoir (in Kelvin) and Tc is the temperature of the cold reservoir (also in Kelvin).
(a) In your case, you are given Th = 50.0°C = 323.15K and Tc = -20.0°C = 253.15K. Plugging these values into the formula:
COP = 323.15K / (323.15K - 253.15K) = 4.62.
Therefore, the best coefficient of performance for this heat pump is 4.62.
(b) To calculate the amount of heat pumped into the warm environment, you need to use the formula:
Qh = W / COP,
where Qh is the amount of heat (in joules) and W is the work input (in joules).
In this case, you are given W = 3.60 x 10^7 J.
Qh = (3.60 x 10^7 J) / 4.62 = 7.81 x 10^6 J.
To convert this to kilocalories, you can use the conversion factor 1 kcal = 4184 J:
Qh = (7.81 x 10^6 J) / 4184 J/kcal ≈ 1862 kcal.
Therefore, the heat pump would pump approximately 1862 kilocalories into the warm environment.
(c) To compare the costs of producing the same amount of heat by the heat pump and burning natural gas, you need to calculate the costs for each method.
Cost of work input = (W / (3.60 x 10^6 J/kW·h)) x (10¢ / kW·h).
In this case, you are given the cost of work input is 10¢/kW·h.
Cost of work input = (3.60 x 10^7 J / (3.60 x 10^6 J/kW·h)) x (10¢ / kW·h) = 100¢.
Cost of natural gas = (Qh / (1.055 x 10^8 J/therm)) x (88.0¢ / therm).
In this case, you are given that a therm equals 1.055 x 10^8 J and the cost of natural gas is 88.0¢ per therm.
Cost of natural gas = (7.81 x 10^6 J / (1.055 x 10^8 J/therm)) x (88.0¢ / therm) = 0.65¢.
Therefore, the cost of producing the same amount of heat by the heat pump (100¢) is higher than the cost of producing the heat by burning natural gas (0.65¢).
(a) The best coefficient of performance (COP) for a heat pump can be calculated using the formula:
COP = 1 / (1 - Tc/Th)
where Tc is the cold reservoir temperature and Th is the hot reservoir temperature.
In this case, the given 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.
Substituting these values into the formula, we get:
COP = 1 / (1 - 253.15/323.15)
COP ≈ 1 / (1 - 0.783)
COP ≈ 1 / 0.217
COP ≈ 4.62
Therefore, the best coefficient of performance for this heat pump is approximately 4.62.
(b) To calculate the amount of heat pumped into the warm environment when a certain amount of work is put into the heat pump, we can use the formula:
Qh = W / COP
where Qh is the heat transferred to the warm environment, W is the work input, and COP is the coefficient of performance.
Given that the work input is 3.60 * 10^7 J (or 10.0 kW·h), and the COP is 4.62 (as calculated earlier), we can substitute these values into the formula:
Qh = (3.60 * 10^7) / 4.62
Qh ≈ 7.79 * 10^6 J
To convert this to kilocalories, we divide by the conversion factor:
Qh ≈ 7.79 * 10^6 / 4184
Qh ≈ 1861 kcal
Therefore, the heat pump would pump approximately 1861 kilocalories of heat into the warm environment when 3.60 * 10^7 J (10.0 kW·h) of work is put into it.
(c) To compare the cost of producing the same amount of heat using the heat pump and the natural gas method, we need to calculate the costs for each option.
The cost of the work input for the heat pump is given as 10¢/kW·h. Since 1 kW·h is equal to 3.60 * 10^7 J, the cost for the work input is:
Cost of work input = 10¢ * (10.0 kW·h) = 100¢ = $1
The cost of the natural gas method can be calculated by converting the given cost per therm to the equivalent cost per joule:
Cost of natural gas = (88.0¢ / (1.055 * 10^8 J)) * (7.79 * 10^6 J)
Cost of natural gas ≈ 68.37¢ ≈ 68¢
Therefore, the cost of producing the same amount of heat by the heat pump method is $1, while the cost of producing the same amount of heat by burning natural gas is 68¢.
To compare these costs, we can calculate the ratio:
Cost of heat pump / Cost of natural gas = $1 / 68¢
Hence, the cost of producing the same amount of heat by the heat pump method is approximately 1.47 times higher than the cost of producing the same amount of heat by burning natural gas.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html
η=1-Tc/Th =1- 253.15/323.15 =0.2166,
η = W/Qh =>
Qh= W/ η =3.6•10⁷/0.2166 = 1.66•10⁸ J= =39648.4 kCal
10 kW•h•10 ¢/kW•h= 100 ¢
(1.66•10⁸/1/055•10⁸)•88 ¢ = 55.05 ¢
100/55.05 = 1.82