Answer:
a. W = 51,194.54 kJ
b. W = 102,390 kJ
c. W = 153,585 kJ
Explanation:
[tex](COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }[/tex]
[tex]W=Q_{1} \frac{T_{1}-T_{2} }{T_{1} }[/tex]
a. the ground at 15°C.
[tex]T_{1}[/tex]=20°C = 273 K + 20 = 293 K
[tex]T_{2}[/tex]=15°C = 273 K + 15 = 288 K
[tex]Q_{1}=3x10^{6} kJ[/tex]
[tex]W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}[/tex]
[tex]W = 0.051195x10^{6} kJ[/tex]
W = 51,194.54 kJ
b. a pond at 10°C.
[tex]T_{2}[/tex]=10°C = 273 K + 10 = 283 K
[tex]W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}[/tex]
[tex]W = 0.102390x10^{6} kJ[/tex]
W = 102,390 kJ
c. the outside air at 5°C.
[tex]T_{2}[/tex]=5°C = 273 K + 5 = 278 K
[tex]W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}[/tex]
[tex]W = 0.153585x10^{6} kJ[/tex]
W = 153,585 kJ
Hope this helps!
(a) For ground at 15°C, the input be "51194 kJ".
(b) For a pond at 10°C, the input be "1.06 × 10⁵ kJ".
(c) For outside air at 5°C, the input be "1.54 × 10⁵ kJ".
According to the question,
Temperature, T₁ = 293 K
T₂ = 288 K
Heat transfer, Q₁ = 3 × 10⁶ kJ
(a) Ground at 15°C
We know that,
→ [tex](COP)_{HP} = (COP)_{Ideal}[/tex]
[tex]\frac{Q_1}{W} = \frac{T_1}{T_1-T_2}[/tex]
or,
The minimum theoretical net work input will be:
→ W = Q₁ × [tex]\frac{T_1-T_2}{T_1}[/tex]
By substituting the values,
= 3 × 10⁶ × [tex]\frac{293-288}{293}[/tex]
= 51194 kJ
(b) A pond at 10°C
Now,
→ W = Q₁ × [tex]\frac{T_1-T_2}{T_1}[/tex]
= 3 × 10⁶ × [tex]\frac{293-283}{293}[/tex]
= 1.06 × 10⁵ kJ
(c) The outside air at 5°C
→ W = Q₁ × [tex]\frac{T_1-T_2}{T_1}[/tex]
= 3 × 10⁶ × [tex]\frac{293-278}{293}[/tex]
= 1.54 × 10⁵ kJ
Thus the responses above are correct.
Find out more information about temperature here:
https://brainly.com/question/19666326
