A) water can be sterilized by boiling. How much heat is required to heat 4 gallons of water from room temperature 72F to 212F (1 g = 3.78 L)  B) if the heat of vaporization of water is 40.7 kJ/mol how much heat is required to convert the original 4 gallons of water not at 212F to steam. Assume the density of water is 1g/mL at 72F

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Chapter6: The States Of Matter
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A) water can be sterilized by boiling. How much heat is required to heat 4 gallons of water from room temperature 72F to 212F (1 g = 3.78 L) 

B) if the heat of vaporization of water is 40.7 kJ/mol how much heat is required to convert the original 4 gallons of water not at 212F to steam. Assume the density of water is 1g/mL at 72F 

Expert Solution
Step 1

Heat of Vaporization:

Heat of vaporization is the heat required to vaporize or evaporate to a certain amount of liquid to its gaseous form. It is represented by Hvap . It is expressed in kJ mol-1

 

To calculate the heat required for any given amount of a substance the equation employed is-

Q = m ×c × Twhere,Q = Heat required (J or kJ)c =specific heat ( Jg°C)T =Difference in temperatureT = Final temperature (Tf) - Initial temperature (Ti)

Step 2

Answer to Part A:

Given,

Volume of water = 4 gallonsInitial Temperature, Ti = 72 °FFinal Temperature, Tf = 212 °FConversion Factor , 1 gallon = 3.78 LDensity of water = 1 g / mL

First, convert gallons of water to Litres and then to millilitres 

1 gallon = 3.78 LThus,4 gallon = 4 ×3.78 L = 15.12 LNow,1 L = 1000 mLThen,15.12 L = 15.12 ×1000 mL = 15120 mLhence,Volume of water , V = 15120 mL

 

Next, use density to find the mass of water

Given, Density = 1 g/mLVolume = 15120 mLDensity = MassVolumeMass = Density ×VolumeMass = 1 gmL ×15120 mLMass of water = 15120 g

Step 3

Convert given temperatures from Fahrenheit to Celsius and find the temperature difference

The formula to find Fahrenheit to Celsius is-(33.8 °F -32) ×59 = 0 °CTi =72 °F (72 °F -32) ×59 = 22.22 °CTi =22.22 °CAlso,Tf =212 °F(212 °F - 32) ×59 =100 °CTf = 100 °CThus, temperature difference,T =(100 - 22.22) °CT =77.78 °C

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