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Energy

ntan athlete ...

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ntan athlete takes 100g of glucose of energy equivalent to 1560 KJ. how much amount of energy is uptaken by 1g molecule of glucosen

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Q. An athlete is given 100 g glucose of energy equivalent to 1560 kJ. He utilizes 50% of this gained energy in an event. Enthalpy of

H_{2} O

for evaporation is

44 k J m o l^{- 1}

. In order to avoid storage of energy in the body, the mass of water (in g) he would need to perspire is _________.

Q. An athlete is given 100 g of glucose

(C_{6} H_{12} O_{6})

of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, the weight of water he would need to perspire is (The enthalpy of evaporation of water is 44 kJ/mol.)

Q. An athlete is given

100 g

(C_{6} H_{12} O_{6})

of energy equivalent to

1560 k J

50 %

of this gained
energy in the event. In order to avoid storage of energy in the body, calculate the weight of water he would need to perspire.
The enthalpy of evaporation of water is

44 k J / m o l

Q. An athelete is given 100 g of glucose

(C_{6} H_{12} O_{6})

of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, Determine the weight of water he would need to perspire. (The enthalpy of evaporation of water is 44 kJ/mole.)

Q. How much energy is gained directly during glycolysis from one molecule of glucose?

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