Question

(b) During one day, 250 kg of water is pumped through

the solar panel. The température of this
water rises from 16°C to 38°C.
The water absorbs 25% of the energy incident on the solar panel. The
specific heat capacity
of water is 4200J/(kg °C).
Calculate the energy incident on the solar panel during that day.
Please explain how to find the incident energy

Answer 1

To find the energy incident on the solar panel, first calculate the energy absorbed by the water (23,100,000 J) using the specific heat capacity formula. Since the water absorbed 25% of the incident energy, divide the energy absorbed by 0.25 to get the total incident energy (92,400,000 J).

Step-by-step explanation:

The student is asking how to calculate the energy incident on the solar panel given that 250 kg of water has absorbed a certain amount of energy (solar heating) as its temperature rose from 16°C to 38°C. The information provided also specifies that the water absorbed 25% of this energy and gives the specific heat capacity of water.

First, we need to calculate the energy absorbed by the water using its specific heat capacity, the mass of the water, and the change in temperature (delta T). The formula for the energy absorbed (Q) is:

Q = mass × specific heat capacity × delta T

Plugging in the values, we get:

Q = 250 kg × 4200 J/(kg°C) × (38°C - 16°C)

Q = 250 kg × 4200 J/(kg°C) × 22°C

Q = 23,100,000 J (or 23.1 MJ)

Since the water absorbed only 25% of the energy incident on the solar panel, we can find the total incident energy by dividing the energy absorbed by the percentage absorbed. Let's denote the total energy incident as E:
E = Q / 0.25

E = 23,100,000 J / 0.25

E = 92,400,000 J

Answer 2

Answer: The energy incident on the solar panel during that day is
`9.24 * 10^(7) J`.

Step-by-step explanation:

Given: Mass = 250 kg

Initial temperature =
`16^(o)C`

Final temperature =
`38^(o)C`

Specific heat capacity = 4200
`J/kg^(o)C`

Formula used to calculate the energy is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = heat energy

m = mass of substance

C = specific heat capacity

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = 250 kg * 4200 J/kg^(o)C * (38 - 16)^(o)C\\= 250 kg * 4200 J/kg^(o)C * 22^(o)C`

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.

`(25)/(100) * q = 250 kg * 4200 J/kg^(o)C * 22^(o)C\\q = 9.24 * 10^(7) J`

Thus, we can conclude that the energy incident on the solar panel during that day is
`9.24 * 10^(7) J`.