Question

The total volume of water on earth is estimated to be 1.386x109 km3. How many kg of glucose (C6H12O6) would need to be combusted to produce enough energy to heat all the water on earth by 1°C? Assume all “earth water” is similar to regular water and has a density of 1 g/cm3 and specific heat capacity of 4.184 kJ/g.K.

Answer 1

We know that the enthalpy of combustion of glucose is 2880 kJ/mol.

Given that volume = 1.386x109 km3.

specific heat capacity of 4.184 kJ/g.K.

temperature change dT = 1°C

1.00 km ^3 = 1.00*10^ 1 5 cm^3

Then ; 1.386x10^9 km3 = 1.386x10^24 cm^3

Now calculate the mass of water = volume / density

= 1.386x10^24 cm^3* 1.0 g * cm^3

= 1.386x10^24 g

Now calculate the amount of energy as follows:Q = mCdT

Here m = mass of water , c= specific heat

dT temperature change

then ;Q= 1.386x 10^24 g *4.184 kJ/g.K*1

= 5.8 10^24 KJ

Then calculate the mole of Glucose as follows:

= total energy ; 5.8 10^24 KJ/2880 kJ/mol

= 2.01*10^21 mole Glucose

Amount of glucose = number of mole * molar mass

= 2.01*10^21 mole Glucose *180.1559 g/mol

= 3.63*10^23 g Glucose

= 3.63*10^20 kg Glucose

Hence

3.63*10^20 kg of glucose (C6H12O6) would need to be combusted to produce enough energy to heat all the water on earth by 1°C