5. A large kiddie swimming pool (width - 10m, length-10m, depth-1 1m) is exposed

Question

5. A large kiddie swimming pool (width - 10m, length-10m, depth-1 1m) is exposed to open sunshine throughout a 10 hour day. At the start of the day the pool water was uniformly warm at A radiometer installed near the pool registered an average total radiation intensity of 300 W/m2. Assume 1/3 of this energy was dissipated in evaporation. Calculate the mass of water evaporated during the 10 hour period. 6. Calculate by how much the depth of water in the pool decreases at the end of the day. 7. If the remaining 2/3 of the energy was spent in warming the water, what will be the temperature of the pool at the end of the 10 hour period?

Explanation / Answer

volume of water =10*10*11 m3 =1100m3

mass of 1 m3 water is 1000kg.(d=mass/vol)

total mass of water= 1100000kg

energy used in evaporation per sec= 1/3*300= 100joules per m2=1joule over the whole pol surface area

total energy used =(3600*10)*1= 36000joules

energy required to evaporate 1 kg of water=2246000joules

total amount of water evaporated is = 36000/2246000=0.16kg

which is equal to=160m3 of water is evaporated.

6.change in depth of water:

let the new height be x

therefore x*10*10=(1100-160)=940

x=9.4m

therefore change in height=11-9.4=1.6 m

7.specific heat capacity of water is =4181J/kg °C

2/3 of total energy=300*2/3 = 200 per m2 = 2 joules per meter

total energy used =(3600*10)*2= 72000joules per

rise in temp =72000/4181= 17.2C

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