On a summer day sunlight may put 100million calories into an80000-liter swimming

Question

On a summer day sunlight may put 100million calories into an80000-liter swimming pool. a) What temperature rise occurs if no other heat entersor leaves the pool? b) How many liters of water would have to evaporate fromthe pool to keep its temperature constant? On a summer day sunlight may put 100million calories into an80000-liter swimming pool. a) What temperature rise occurs if no other heat entersor leaves the pool? b) How many liters of water would have to evaporate fromthe pool to keep its temperature constant?

Explanation / Answer

For 1000 liters = 1 cubic meter and 1 calorie = 4.184 J, thespecific heat Cp of water varies slightly with temperature (from4.1784 at 30 degr. to 4.21 or more when colder or hotter) Initial temperature not provided, average Cp at 20 degrees isCp = 4.1818 J/(g.degreeK), almost 1 calorie per gram of water perdegree (you could use that approximation). If you want to useJoules, 100 million calories are 418,400,000 J. How many grams ofwater in 80,000 liters, i.e. 80 cubic meter? The density of waterchanges slightly with temperature. At 20 degrees it is 998.2071kg/cubic meter. Therefore the 80 cubic meters have (80)(998.2071)kgof water, i.e. 79,856.569 kg. But the Cp quantity is in grams, sowe need the water in grams, multiply kg by 1000: 79,856,569.00g. (a) Dividing the Joules of heat we get from the amount ofwater we have, this heat will raise the water temperature 5.239 393709 degrees Celsius or Kelvin, i.e. 5.24 degrees. (b) While water is evaporating it uses heat but keeps thetemperature constant until significant evaporation stop. Wewould need to use all the heat applied here to evaporate waterinstead of warming it up. The heat of partial vaporization ofsurface water when the ambient temperature is near 25 degreesis 43990 J for every 18 g (1 mol) of water. Dividing the heatapplied(418,400,000 J0 by 43990 gives us 9511.252 557 sets of18 g (multiply 18 by 9511.252557). We get 171,202.546 g of waterthat have to evaporate, that is 171.202546 kg of water. We havekept the temperature constant at between 20 to 22 degrees justabout. Density of water is still that of 20 degrees: 998.2071kg in every 1000 liters, i.e. 0.9982071 kg per liter. Dividing171.202546 kg by 0.9982071 kg/liter we get 171.510 046 4 liters˜ 171.5 liters. ANSWERS: (a) 5.24 degrees (b) 171.5 liters

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