Question

a 280 kg black bear hibernation the bear's body temperature drops from 37 to 33 degrees celsius. how many grams of glucose must the bear metabolize in order to restore its body temperature to normal? assume the bear's body is mostly water with a specific heat capacity of 4.2 J/(g. degrees celsius) and that all the energy from the combustion of glucose is used to raise the bear's body temperature. (Glucose:DH(combustion)=-2830 kj/mol, molar mass=180 g/mol)

Answer 1

mass of bear = 280 kg or 280,000 g

heat capacity, =4.2 J/(g. degrees celsius)

temperature   change = 37-33 degree= 4 C

Glucose:DH(combustion)=-2830 kj/mol,

molar mass of glucose =180 g/mol

first calculate the amount of heat as follows:

mass* specific heat

= 280,000 g * 4.2 J/(g. degrees Celsius ) * 4 degrees Celsius

= 4704000 J

= 4704 KJ

Now calculate the number of mole of glucose as follows:

Total heat /   combustion of glucose eper mole

= 4704 KJ/ 2830 KJ/ mole

= 1.66 mole Glucose

Now calculate the mount of glucose as follows:

Number of moles * molar mass

= 1.66 mole Glucose * 180 g/ mole

= 299.19 g

= 300 g glucose

Hence 300 g glucose must the bear metabolize in order to restore its body temperature to normal