Question

A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2.

What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2 -area collector? (cw = 4 186 J/kg⋅°C)

a. 14.3°Cb. 22.4°Cc. 28.7°Cd. 44.3°C

Answer 1

c. 28.7C

Step-by-step explanation:

Since the solar radiation incident on the panel is 1000W/m2 and the collector has an area of 4m2. We can conclude that the solar power generated is

1000 * 4 = 4000 W or J/s

Within 1 hour, or 3600 seconds this solar power generator should generate (with 25 % efficiency):

E = 0.25 * 4000 * 3600 = 3600000 J or 3.6 MJ

This energy will be converted to heat to heat up water. Using water heat specific of 4186 J/kg C we can find out how much temperature has raised:

`\Delta T = (E)/(c*m) = (3600000)/(30 * 4186) = 28.7^oC`

So C. is the correct answer