Answer: the total amount of energy transferred by work into the heat pump over the 14-day period is 5.8GJ
Explanation: since the meter reading for the 14 days period is 1620kW.h
Electrical power Ep is
Ep = 1620*10^3W.h
In 14 days there are 14*24 hrs = 336hrs
Ep = (1620*10^3)/336 = 4821.42W
This means 4821.42J of work is done per sec. Therefore for 14 days we have,
3600*24*14*4821.42J of enegy
= 5831989632J = 5.8GJ
Answer 2 = the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period is 41.2GJ
Explanation: rate of energy transfer as heat to building is 1.4*10^5 kJ/hr
That is 1.4*10^8 J/hr
In 14 days it becomes
24*14*1.4*10^8 = 47GJ
Energy received from well by heat pump =
Energy transfered - energy received
= 47 - 5.8 =41.2GJ
Answer 3. the heat pump’s coefficient of performance is 0.12
Explanation: COP = energy obtained/ work done
COP = 5.8/47 = 0.12
In this exercise we have to use knowledge about forces to determine the heat coefficient, so we can say that:
The total amount of energy transferred is 5.8GJ
Knowing that the potential energy formula is given by:
[tex]Ep = 1620*10^3W.h\\ 14*24 hrs = 336hrs\\ Ep = (1620*10^3)/336 = 4821.42W\\ 600*24*14*4821.42J \\ = 5831989632J = 5.8GJ [/tex]
This energy that can be transferred within 14 days, we will have that corresponds to:
[tex]24*14*1.4*10^8 = 47GJ[/tex]
Then the formula can be written as follows:
[tex]Energy \ received \ heat \ pump = Energy \ transfered - energy \ received \\ = 47 - 5.8 =41.2GJ[/tex]
The coefficient will be calculated as:
[tex]COP = energy \ obtained/ work \ done]\\ COP = 5.8/47 = 0.12[/tex]
See more about coefficient at brainly.com/question/11808898
