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Sagot :
Answer: The energy incident on the solar panel during that day is [tex]9.24 \times 10^{7} J[/tex].
Explanation:
Given: Mass = 250 kg
Initial temperature = [tex]16^{o}C[/tex]
Final temperature = [tex]38^{o}C[/tex]
Specific heat capacity = 4200 [tex]J/kg^{o}C[/tex]
Formula used to calculate the energy is as follows.
[tex]q = m \times C \times (T_{2} - T_{1})[/tex]
where,
q = heat energy
m = mass of substance
C = specific heat capacity
[tex]T_{1}[/tex] = initial temperature
[tex]T_{2}[/tex] = final temperature
Substitute the values into above formula as follows.
[tex]q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C[/tex]
As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.
[tex]\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J[/tex]
Thus, we can conclude that the energy incident on the solar panel during that day is [tex]9.24 \times 10^{7} J[/tex].
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