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? (c_w = 4 186 J/kg*$\degree$C) A)14.3 $\degree$C B)22.4 $\degree$C C)28.7 $\degree$C D)44.3 $\degree$C

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of energy absorbed by the collector in one hour:

Energy absorbed = (solar radiation incident)*(collector area)*(conversion efficiency)*(time)
Energy absorbed = (1000 W/m^2) * (4.00 m^2) * (0.250) * (3600 s)
Energy absorbed = 3,600,000 J

Next, we need to use this energy to calculate the temperature increase of the water:

Energy absorbed = (mass of water)*(specific heat capacity of water)*(temperature increase)
Temperature increase = Energy absorbed / (mass of water * specific heat capacity of water)
Temperature increase = 3,600,000 J / (30.0 kg * 4,186 J/kg*C)
Temperature increase = 28.7 C

Therefore, the temperature of the water will increase by 28.7 degrees Celsius, or Choice C.

A Solar Heating System Has a 25 °\degree° C) A)14 °\degree° C B)22

A Solar Heating System Has a 25 $\degree$ C)
A)14 $\degree$ C
B)22