Question

James Watt once observed that a hard-working horse can lift a 330 lb weight 100. ft in 1 minute. Assuming the horse generates energy to accomplis this work by oxidatively metabolizing glucose

C6H12O6(s) + 6 O2(g)----> CO2(g) + 6 H2O(l)

ΔGo = -2880 kJ and that only 40% of the chemical energy can be converted to work, calculate how much glucose the horse must metabolize to sustain this rate of work for one hour at 298 K.

Answer 1

Given data

ΔGo = -2880 kJ

Work generated by chemical energy = 0.4 * (-2880) = - 1152 kJ

Work performed by the horse

W = mass x gravity constant x height

=( 330 lb x 0.4536 kg/lb) x (9.81 m/s2) x (100 ft x 0.3048 m/ft) x ( 1 min-1 x 60min/hr)

= 2685481 J/hr

= 2685.481 kJ/hr

Amount of glucose that the horse requires

= (Work performed by the horse) x (Molecular weight of glucose) /( work generated by chemical energy)

= 2685.481 * 180.18/1152

= 420 g of glucose the horse must metabolize to sustain this rate of work for one hour at 298 K.