How many grams of water can be cooled from 35 degrees celcius to 22 degrees by the evaporation of 50 grams of water? Assume heat of vaporization of water is 2.4 kj/g. The specific heat of water is 4.18 j/g*degrees celcius. Please show your work.
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First off, note the mistake in the first answer. The final temperature is 22 (not 35) so this will give you the opposite of what you want. it should be (22-35)
Anyways, This is a basic example of an energy balance. Energy has to be conserved. In other words, the energy that was added to the water in order to boil it had to have come from cooling this other water. The change in enthalpy has to be zero since we cannot lose or create energy. The amount of energy consumed in evaporating the water is simply the amount of water boiled times the heat of vaporization.
dH=m*L. That is 50*2.4=120kJ. Note that these units are kJ (heats of vaporization are much larger than heat capacities or specific heats)
Moving on, this change in enthalpy must be the amount of energy lost from the water cooling to 22 degrees. The amount of energy lost to a temperature change can be roughly approximated to mc(deltaT). since the total change in enthalpy has to be equal to zero, 120+mcdT=0
mcdT=-120kJ
we know the change in temperature and the heat capacity so we solve for the amount of water, m.
-120/.00418/(22-35)=m m=2208.32 grams of water.
You might be wondering why I used .00418 instead of 4.18. Remember that the units for the heat of vaporization were kj/g while the units for the heat capacity are j/g so we have to have consistent units. I'm trying to become a chemical engineer and unit issues always bite you in ***!
Also try to appreciate the amount of water that was cooled. Vaporizing 50 grams cooled 2208.32 grams! Vaporizing something results in a huge enthalpy change! Thats why we use steam and water to heat and cool so many the processes we do every day.
Anyways, This is a basic example of an energy balance. Energy has to be conserved. In other words, the energy that was added to the water in order to boil it had to have come from cooling this other water. The change in enthalpy has to be zero since we cannot lose or create energy. The amount of energy consumed in evaporating the water is simply the amount of water boiled times the heat of vaporization.
dH=m*L. That is 50*2.4=120kJ. Note that these units are kJ (heats of vaporization are much larger than heat capacities or specific heats)
Moving on, this change in enthalpy must be the amount of energy lost from the water cooling to 22 degrees. The amount of energy lost to a temperature change can be roughly approximated to mc(deltaT). since the total change in enthalpy has to be equal to zero, 120+mcdT=0
mcdT=-120kJ
we know the change in temperature and the heat capacity so we solve for the amount of water, m.
-120/.00418/(22-35)=m m=2208.32 grams of water.
You might be wondering why I used .00418 instead of 4.18. Remember that the units for the heat of vaporization were kj/g while the units for the heat capacity are j/g so we have to have consistent units. I'm trying to become a chemical engineer and unit issues always bite you in ***!
Also try to appreciate the amount of water that was cooled. Vaporizing 50 grams cooled 2208.32 grams! Vaporizing something results in a huge enthalpy change! Thats why we use steam and water to heat and cool so many the processes we do every day.