The incident power of the solar radiation is 1,185.6 W.
The thermal energy gained by the water is calculated as follows;
Q = mcΔθ
Q = 0.019 x 4200 x (72 - 20)
Q = 4,149.6 J
The incident power of the solar radiation is calculated as follows;
[tex]0.7\times P = \frac{Q}{t} \\\\P = \frac{Q}{0.7t} \\\\P = \frac{4149.6}{0.7 \times 5} \\\\P = 1,185.6 \ W[/tex]
Thus, the incident power of the solar radiation is 1,185.6 W.
The complete question is below
In 5.0 s, 0.019 kg of water flows through the tubes of a solar heater. The temperature of the water increases from 20 °C to 72 °C. The specific heat capacity of water is 4200 J / (kg °C). Calculate the thermal energy gained by the water in 5.0 s.
The efficiency of the solar panel is 70%. Calculate the power of the solar radiation incident on the panel.
Learn more about incident power of solar radiation here: https://brainly.com/question/10014938
