Calculations of how much water evaporated and what is left

In summary, the conversation discusses the process of heating water on a stove and determining how much water is left after adding a specific amount of heat. The heat of fusion and vaporization for water are also mentioned. The solution involves calculating the amount of heat used to heat the water, then using a ratio to determine how much water can be evaporated with the remaining heat, and finally subtracting the evaporated amount from the initial amount of water to find the remaining amount.
  • #1
apbuiii
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0

Homework Statement


Suppose you put a coffee pot containing 300 g of water at 20 C on the stove. You then add 350 kJ of heat to the water, which heats the water to 100 C and the water starts to boil. After all the heat is added to the water, then how much water do you still have left? The heat of fusion for water is 6.01 kJ/mol and the heat of vaporization is 40.7 kJ/mol.


Homework Equations



q=mass(specific heat)(Tf-Ti); q=moles(delta-enthalpy)

The Attempt at a Solution


I first found how much energy it would take to evaporate all of the 300 grams. (100-20)(300g)(4.18) + (40.7 KJ)(300)(1mol/18g)(1000J)= 778653 J. So I then set up a ratio that if 778653 evaporates 300 grams then how many grams will 350000 J evaporate. 778653/300 = 350000/X. That gave me 135 grams evaporated. So I then minused that from the 300 giving me 165 grams left. Seemed reasonable to me but I got it wrong! Please help, but showing the steps to find the answer would be even more helpful! Thanks
 
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  • #2
There are two heat sinks - one (water heating to 100 °C) doesn't depend on the amount of evaporated water, other (water evaporation) depends. When you use ratio it is equivalent to assumption that you have heated only part of water, while you have to heat it all.

Calculate how much heat was used to heat the water, excess heat was consumed by evaporation.
 
  • #3
Okay, so I did (300)(4.18)(100-20)= 100320 J. Then I subtracted that from 350000J to get 249680 J. This is where I get a little confused... Do I then do this: 249680=mols(40.7KJ)(1000J). That gave me 6.13 moles of water and then I change that to grams to give me 110.4 grams of water that it can evaporate. I then subtract 300-110.4 to give me the grams of water left 190 grams. Is that correct? Thank you so much...
 
  • #4
I have not checked the numbers, but the logic is OK.
 
  • #5
!

As a scientist, my response to this content would be to first acknowledge that the student has attempted to use the correct equations and has shown their work, which is a good start. However, there are a few errors in their calculations that may have led to the incorrect answer.

First, the student used the specific heat of water (4.18 J/g°C) instead of the specific heat of vaporization (40.7 kJ/mol) in their first calculation. This would result in a significantly lower amount of energy needed to evaporate the water. The correct equation for this step would be q=m(Hvap), where q is the energy needed, m is the mass of water, and Hvap is the specific heat of vaporization.

Next, in their ratio calculation, the student used the incorrect conversion factor for moles to grams. The correct conversion factor would be 1 mol/18 g, not 1000 J. This would result in a significantly higher amount of water evaporated, and therefore a lower amount of water remaining.

Taking these errors into account, the correct calculation would be: (100-20)(300g)(40.7 kJ/mol) + (40.7 kJ)(300g)(1mol/18g) = 1.56 x 10^7 J. Using the same ratio calculation, we find that this amount of energy would evaporate approximately 378 grams of water. Therefore, the amount of water remaining would be 300g - 378g = -78g, which does not make sense. This could indicate that there is an error in the given information or that the student may have made a mistake in their calculations.

In conclusion, as a scientist, I would suggest checking the given information and reviewing the calculations to determine the correct amount of water remaining after adding 350 kJ of heat to the coffee pot. Additionally, it would be beneficial to double-check all conversions and equations used to ensure accuracy in the calculations.
 

What is the purpose of calculating how much water evaporated and what is left?

The purpose of this calculation is to understand the water cycle and how water moves through the environment. It can also help determine the amount of water available for human use, such as for agriculture or drinking.

How is the calculation of water evaporation and remaining water done?

The calculation is typically done using data on temperature, humidity, wind speed, and other environmental factors. These variables are used in equations to estimate the rate of evaporation and the amount of water remaining. Advanced techniques such as remote sensing can also be used to track changes in water levels over time.

What factors affect the amount of water that evaporates?

The amount of water that evaporates is influenced by several factors, including temperature, humidity, air pressure, wind speed, and the surface area of the water. Other factors such as the presence of impurities in the water or the type of surface the water is on can also impact evaporation rates.

Why is it important to track water evaporation and remaining water?

Tracking water evaporation and remaining water is important for understanding and managing our water resources. It can help us plan for droughts and floods, manage water supplies for agriculture and industry, and monitor changes in the environment caused by climate change.

How accurate are these calculations of water evaporation and remaining water?

The accuracy of the calculations depends on the quality of the data and the methods used. In general, the calculations can give a good estimate of the evaporation rate and remaining water, but there can be some variability due to the complexity of the water cycle and the difficulty in measuring all of the relevant factors. Ongoing research and advancements in technology are continuously improving the accuracy of these calculations.

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