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 4200 J / (kg °C). Calculate the energy incident on the solar panel during that day. *

Answer 1

Answer:

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

Answer 2

Answer:

92.4kg/j

Explanation:

mass=250kg

temperature change=38-16=22

mass*temperature change =250*1200*22

this is 25% pf the total energy fall

25/100=250*4200*22

x=4*250*4200*22

x=9.24*10=92.4kg/J