 
    
 To get work in Joules (the SI unit of energy), we must transform
	the external pressure to Pascals and
 
	To get work in Joules (the SI unit of energy), we must transform
	the external pressure to Pascals and   to cubic meters:
  to cubic meters:
	  
 
  
 
 C.  1.5L is
 C.  1.5L is
	  .
	Using the density of water at
 .
	Using the density of water at   C, we
	find
 C, we
	find
	  
 
The heat required to bring the water to its boiling point is therefore
  
 
This heat is supplied by a 1500W heating element. A Watt is a Joule per second so
  
 
	  
 
 of heat, i.e. the water gains
		this much heat.
  of heat, i.e. the water gains
		this much heat. C to
 C to   C.
		Relative to the lead spheres, this process yields
 C.
		Relative to the lead spheres, this process yields
		  of heat.  Thus the water gains
  
		of heat.  Thus the water gains   .
 .
	 .
	The heat gained by the water in going from 20 to
 .
	The heat gained by the water in going from 20 to   C is
 C is
	  
 
Therefore
  
 
	  
 
 .  There are 24 boxes so
	the total work done is 3.5kJ.
	Relative to the person doing the lifting,
 .  There are 24 boxes so
	the total work done is 3.5kJ.
	Relative to the person doing the lifting,   .
	To compensate for this energy use, the person must take in at
	least
	3.5kJ of usable (catabolizable) energy in the form of food.
	Carbohydrates provide 4.0kcal/g which is 16.7kJ/g.
	Therefore
 .
	To compensate for this energy use, the person must take in at
	least
	3.5kJ of usable (catabolizable) energy in the form of food.
	Carbohydrates provide 4.0kcal/g which is 16.7kJ/g.
	Therefore
	  
 
Note that this calculation assumes 100% efficiency of all the processes involved (muscle work, extraction of nutrients from food, etc.). Nevertheless, it's safe to conclude that it won't be necessary to eat double portions.
	  
 
  
 
For this reaction,
  
 
	  
 
 .
		Therefore
 .
		Therefore   For ideal gases, PV=nRT while for solids and liquids,
		the product PV is insignificantly small.  Accordingly,
 
		For ideal gases, PV=nRT while for solids and liquids,
		the product PV is insignificantly small.  Accordingly,
		  
 
		where   is the change in the number of moles of
		gas during the reaction.  For this reaction (as written),
		there are
  is the change in the number of moles of
		gas during the reaction.  For this reaction (as written),
		there are
		  reactant molecules and 2 product molecules
		so
  reactant molecules and 2 product molecules
		so   .  That gives
 .  That gives
		
  
 
	  
 
