Answer:
Part a)
[tex]Q_{rem} = 1400 kW[/tex]
Part b)
[tex]\frac{dm}{dt} = 83.6 kg/s[/tex]
Explanation:
Part a)
Energy delivered to the building is given as
[tex]E_{out} = 2000 kW[/tex]
Work done by the electricity is given as
[tex]W = 600 kW[/tex]
so we have
[tex]Q_{rem} + W = E_{out}[/tex]
[tex]Q_{rem} = 2000 - 600[/tex]
[tex]Q_{rem} = 1400 kW[/tex]
Part b)
As we know that water in and out temperature is given as
[tex]T_1 = 7 degree[/tex]
[tex]T_2 = 3 degree[/tex]
so we have
[tex]Q = \frac{dm}{dt} s \Delta T[/tex]
[tex]1400 \times 10^3 = \frac{dm}{dt} (4186) (7 - 3)[/tex]
[tex]\frac{dm}{dt} = 83.6 kg/s[/tex]
This question involves the concepts of specific heat capacity and the law of conservation of energy.
(a) "1400 KW" energy is removed from water each second.
(b) "83.33 kg/s" water moves through the system each second.
(a)
Applying the law of conservation of energy to this situation we get:
[tex]E_r + W=E_d[/tex]
where,
[tex]E_r[/tex] = Energy removed from water per second = ?
W = Electrical Power Consumed = 600 KW
[tex]E_d[/tex] = Energy delivered to buildingper second = 2000 KW
Therefore,
[tex]E_r+600\ KW=2000\ KW\\E_r=2000\ KW-600\ KW\\E_r=1400\ KW[/tex]
(b)
Now, we will use the formula of specific heat capacity to find out the mass flow rate of water:
[tex]E_r=\frac{m}{t}C\Delta T[/tex]
where,
m/t = mass flow rate of water = ?
C = Specific Heat Capacity of water = 4200 J/kg.°C
ΔT = Change in temperature = 7°C - 3°C = 4°C
Therefore,
[tex]1400\ x\ 10^3\ W=\frac{m}{t}(4200\ J/kg.^oC)(4\ ^oC)\\\\\frac{m}{t}=\frac{1400\ x\ 10^3\ W}{(4200\ J/kg.^oC)(4\ ^oC)}\\\\\frac{m}{t}=83.33\ kg/s[/tex]
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
