Question

(b) During one day, 250 kg of water is pumped through

Answer 1

Answer: The energy incident on the solar panel during that day is $9.24 \times 10^{7} J$.

Explanation:

Given: Mass = 250 kg

Initial temperature = $16^{o}C$

Final temperature = $38^{o}C$

Specific heat capacity = 4200 $J/kg^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass of substance

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C$

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.

$\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J$

Thus, we can conclude that the energy incident on the solar panel during that day is $9.24 \times 10^{7} J$.