A solar heating system has a 30.0% conversion efficiency;the solar radiation incident on the panels is 900 W/m².What is the increase in temperature of 32.0 kg of water in a 1.00-h period by a 5.00-m²-area collector? (c_w = 4 186 J/kg⋅°C)

Question

Quizplus · Accepted Answer

First, we need to calculate the total energy incident on the collector in one hour:

E = (900 W/m²) x (5.00 m²) x (3600 s) = 1.62 x 10^7 J

Then, we need to calculate the energy absorbed by the water. Since the conversion efficiency is 30%, only 30% of the incident energy is absorbed:

E_abs = (0.30) x (1.62 x 10^7 J) = 4.86 x 10^6 J

Finally, we can use the specific heat of water to calculate the increase in temperature:

ΔT = E_abs / (m x c)

where m is the mass of water and c is the specific heat of water. Plugging in the given values:

ΔT = (4.86 x 10^6 J) / (32.0 kg x 4 186 J/kg⋅°C) = 36.3°C

Therefore, the increase in temperature of the water is 36.3°C in one hour, which corresponds to choice C.

A Solar Heating System Has a 30