A water body of 2000m^3 and 2m depth has a absorption of 100J/m^2/sec. Assuming no energy loss, how much would the temperature of the water increase in 2 hours? ?

First, we need to calculate the total energy absorbed by the water body in 2 hours:

Total energy absorbed = absorption rate x surface area x time
Total energy absorbed = 100 J/m^2/sec x 2000 m^2 x 2 hours x 3600 sec/hour
Total energy absorbed = 100 J/m^2/sec x 2000 m^2 x 2 x 3600 sec
Total energy absorbed = 100 J/m^2/sec x 2000 m^2 x 7200 sec
Total energy absorbed = 144,000,000 J

Next, we need to calculate the change in temperature of the water body using the specific heat capacity of water, which is 4186 J/kg°C:

Change in temperature = Total energy absorbed / (mass of water x specific heat capacity)
Change in temperature = 144,000,000 J / (2000 m^3 x 1000 kg/m^3 x 4186 J/kg°C)
Change in temperature = 144,000,000 J / (2,000,000 kg x 4186 J/kg°C)
Change in temperature = 144,000,000 J / 8,372,000 J/°C
Change in temperature = 17.2°C

Therefore, the temperature of the water body would increase by 17.2°C in 2 hours.

1. Calculate the total energy absorbed:

Energy absorbed = Absorption rate x Time x Area

Absorption rate = 100 J/m^2/sec
Time = 2 hours = 7200 sec
Area = 2000 m^3 / 2 m = 1000 m^2

Energy absorbed = 100 J/m^2/sec x 7200 sec x 1000 m^2
= 7.2 x 10^8 J

2. Calculate the change in temperature:

Change in temperature = Energy absorbed / (Mass of water x Specific heat capacity of water)

Mass of water = Volume x Density
Density of water = 1000 kg/m^3

Mass of water = 2000 m^3 x 1000 kg/m^3
= 2 x 10^6 kg

Specific heat capacity of water = 4187 J/kg-K

Change in temperature = 7.2 x 10^8 J / (2 x 10^6 kg x 4187 J/kg-K)
= 8.56 K

Therefore, the temperature of the water would increase by 8.56 K in 2 hours.