Chapter 10, Problem 10.137QP

Both glucose and fructose arc simple sugars with the same molecular formula of C₆H₁₂O₆. Sucrose (C₁₂11₂₂O₁₁), or table sugar, consists of a glucose molecule bonded to a fructose molecule (a water molecule is eliminated in the formation of sucrose), (a) Calculate the energy released when a 2.0-g glucose tablet is burned in air. (b) To what height can a 65-kg person climb after ingesting such a tablet, assuming only 30% of the energy released is available for work. (See the hint for Problem 10.127.) Repeat the calculations for a 2.0-g sucrose tablet.

Interpretation Introduction

Interpretation:

The energy changes in 2.0 g of Glucose and Sucrose tablet has to be calculated and the height the person can climb after ingesting the tablet has to be calculated.

Concept Introduction:

Heat is flow of thermal energy involving two bodies at various temperatures.  The flow of thermal energy involving two bodies at various temperatures is called as heat.  The flow of heat implies that energy is either released or absorbed on describing about energy changes that takes place during a process.

Exothermic process is the chemical process in which heat is released to the surroundings.

Endothermic process is the chemical process in which heat is absorbed from the surroundings.

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation ( ${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

To calculate: The energy changes for 2 grams of Glucose and Sucrose

Answer to Problem 10.137QP

The heat change for Glucose is $\text{31}\text{kJ}$

The heat change for Sucrose is $\text{33}\text{kJ}$

Explanation of Solution

Weight of Glucose = $\text{2}\text{.0}\text{g}$

Molar mass of Glucose = $\text{180}\text{.2}\text{g}$

Heat of combustion of Glucose = $\text{-2801}\text{kJ}{\text{mol}}^{\text{-1}}$

Moles of Glucose = $\text{2}\text{.0}\text{g×}\frac{\text{1}\text{mol}\text{Glucose}}{\text{180}\text{.2}\text{g}\text{Glucose}}$

= $\text{0}\text{.011}\text{mol}$

Moles of Sucrose = $\text{2}\text{.0}\text{g×}\frac{\text{1}\text{mol}}{\text{324}\text{.3}\text{g}}\text{=0}\text{.0058}\text{mol}$

Moles of Glucose = $\text{0}\text{.011}\text{mol}$

Moles of Sucrose = $\text{0}\text{.0058}\text{mol}$

To calculate the energy changes for 2 grams of Glucose and Sucrose

Energy change in Glucose = $\text{0}\text{.011}\text{mol×}\frac{\text{2801}\text{kJ}}{\text{1}\text{mol}\text{}\text{Glucose}}$

= $\text{31}\text{kJ}$

Energy change in Sucrose = $\text{0}\text{.0058}\text{mol×}\frac{\text{5644}\text{kJ}}{\text{1}\text{mol}}$

= $\text{33}\text{kJ}$

Energy change in Glucose = $\text{31}\text{kJ}$

Energy change in Sucrose = $\text{33}\text{kJ}$

Conclusion

The energy changes for 2 grams of Glucose and Sucrose were calculated using the values the moles and heat of combustions of Glucose and Sucrose.  The energy changes for 2 grams of Glucose and Sucrose were found to be $\text{31}\text{kJ}$ and $\text{33kJ}$.

Interpretation Introduction

Concept Introduction:

The energy possessed by an object due to its relative point to some other object, stress surrounded by itself, its electric charge or any other external factor is called as potential energy.

Potential energy can be calculated using the formula,

$\text{PE=mgh}$

Answer to Problem 10.137QP

Height climbed by the person on ingesting Glucose is 15 m.

Height climbed by the person on ingesting Sucrose is 16 m.

Explanation of Solution

To calculate the heat in Joules

$\begin{array}{l}\text{Heat of Glucose=31}\text{kJ×0}\text{.30=9}\text{.3}\text{kJ=9}{\text{.3×10}}^{\text{3}}\text{J}\\ \\ \text{Heat of Sucrose=33}\text{kJ×0}\text{.30=9}\text{.9kJ=9}{\text{.9×10}}^{\text{3}}\text{J}\end{array}$

Heat of Sucrose = 9.9×10³ J

Heat of Glucose = 9.3×10³ J

To calculate the height climbed

$\begin{array}{l}\text{Height}\text{climbed}\text{by}\text{person(Injecting}\text{Glucose)}\\ \\ =\text{9}{\text{.3×10}}^{\text{3}}\text{J=65kg×9}\text{.8}{\text{ms}}^{\text{-2}}\text{×h}\\ \\ \text{h=15}\text{m}\end{array}$

$\begin{array}{l}\text{Height}\text{climbed}\text{by}\text{person(Injecting}\text{Sucrose)}\\ \\ \text{=9}{\text{.9×10}}^{\text{3}}\text{J=65kg×9}\text{.8}{\text{ms}}^{\text{-2}}\text{×h}\\ \\ \text{h=16}\text{m}\end{array}$

Height climbed by the person on ingesting Glucose =15 m.

Height climbed by the person on ingesting Sucrose =16 m.

Conclusion

Heights climbed by the person on ingesting Glucose and Sucrose were calculated using the values of mass, acceleration due to gravity and heat.  The heights climbed by the person on ingesting Glucose and Sucrose were found to be 15 m and16 m.