A55-kg hiker starts at an elevation of 1600 m and climbs to the top of a 3100-m peak. (a) what is the hiker's change in potential energy? (b) what is the minimum work required of the hiker? (c) can the actual work done be more than this? explain why.

Answer

given,

mass of hiker = 55 Kg

elevation at start, h₁ = 1600 m

elevation at top , h₂ = 3100 m

a) change in potential energy

             ΔP = 808500 J

b) The minimum work required by the hiker will be equal to the change in potential energy.

The minimum work done

  W = Δ PE

      = 808500 J

c) Yes, work done can be more than this because hiker will exert more force to balance the force due to gravity.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂