Question

James Watt once observed that a hard-working horse can lift a 330 lb mass 100 ft in 1 min. Assuming the horse generates energy to accomplish this work by metabolizing glucose: C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l) How much (in gram) glucose must a horse metabolize to sustain this rate of work for 1 hour, if the temperature is 250C250C? (1 lb = 0.4536 kg, 1 ft = 0.305 m, g=9.81 ms2ms2)

Answer 1

Ans. # Step 1: Work done by the horse during lifting the mass, W = m g h

            Where, m = mass in kg        ; h = height in m       g = 9.81 m s^-2

            Or, W = (330 lb x 0.4536 kg lb^-1) x 9.81 m s^-2 x (100 ft x 0.305 m ft^-1)

            Hence, W = 44787.39804 kg m² s^-2

            Hence, W = 44787.39804 J                                               ; [1 kg m² s^-2 = 1 J]

                        W = 44.78739804 kJ

# Step 2: At 25.0⁰C, the free energy change of ATP hydrolysis = -30.5 kJ/mol

So, moles of ATP required = Work done / Molar enthalpy of ATP hydrolysis

                                                = 44.78739804 kJ / (30.5 kJ/mol)

                                                = 1.46843928 mol

Hence, required amount of ATP = 1.46843928 mol ATP

# Step 3: 1 mol glucose produces (net gain) 30 mol ATP upon complete oxidation (cytoplasmic NADH produces 1.5 ATP, whereas mitochondrial NADH produces 2.5 ATP. If cytoplasmic NADH is considered to produce 2.5 ATP, net ATP gain would be 32 ATP).

That is, 1 mol glucose produced 30 mol ATP.

So,

            Required amount of glucose = Required moles of ATP / ATP yield of glucose

                                                            = 1.46843928 mol ATP / (30 mol ATP / mol glucose)

                                                            = 0.048947976 mol glucose

Now,

            Requires mass of glucose = Required moles of glucose x Molar mass

                                                            = 0.048947976 mol x (180.1559 g/ mol)

                                                            = 8.818 g

Therefore, required mass of glucose = 8.818 g