Question

(b) During one day, 250 kg of water is pumped through the solar panel. The temperature of this
water rises from 16°C to 38°C.
The water absorbs 25% of the energy incident on the solar panel. The specific heat capacity
of water is 4200J /(kg °C).
Calculate the energy incident on the solar panel during that day.

please exaplain it.

Answer 1

Answer:

Explanation:

Q (heat absorbed) = C * M * (T2 - T1)

Q = 4200 J / (kg deg C) * 250 kg * 22 deg C

Q = 2.31 * 10E7 J    heat absorbed by water

So the total incident energy = 4 Q   since efficiency is 25%

E (total incident energy) = 4 * 2.31 * 10E7 J = 9.24 * 10E7 J