Citric Acid is the correct answer because it contains a density of 1.66 g/cm3, whereas water= 1.00 g/cm3, Olive oil= 0.93 g/cm3, Ethyl alcohol= 0.79 g/cm3
Answer: The energy incident on the solar panel during that day is
.
Explanation:
Given: Mass = 250 kg
Initial temperature = ![16^{o}C](https://tex.z-dn.net/?f=16%5E%7Bo%7DC)
Final temperature = ![38^{o}C](https://tex.z-dn.net/?f=38%5E%7Bo%7DC)
Specific heat capacity = 4200 ![J/kg^{o}C](https://tex.z-dn.net/?f=J%2Fkg%5E%7Bo%7DC)
Formula used to calculate the energy is as follows.
![q = m \times C \times (T_{2} - T_{1})](https://tex.z-dn.net/?f=q%20%3D%20m%20%5Ctimes%20C%20%5Ctimes%20%28T_%7B2%7D%20-%20T_%7B1%7D%29)
where,
q = heat energy
m = mass of substance
C = specific heat capacity
= initial temperature
= final temperature
Substitute the values into above formula as follows.
![q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C](https://tex.z-dn.net/?f=q%20%3D%20250%20kg%20%5Ctimes%204200%20J%2Fkg%5E%7Bo%7DC%20%5Ctimes%20%2838%20-%2016%29%5E%7Bo%7DC%5C%5C%3D%20250%20kg%20%5Ctimes%204200%20J%2Fkg%5E%7Bo%7DC%20%5Ctimes%2022%5E%7Bo%7DC)
As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.
![\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J](https://tex.z-dn.net/?f=%5Cfrac%7B25%7D%7B100%7D%20%5Ctimes%20q%20%3D%20250%20kg%20%5Ctimes%204200%20J%2Fkg%5E%7Bo%7DC%20%5Ctimes%2022%5E%7Bo%7DC%5C%5Cq%20%3D%209.24%20%5Ctimes%2010%5E%7B7%7D%20J)
Thus, we can conclude that the energy incident on the solar panel during that day is
.
Answer:
answer is DE I hope it will help you please follow me