Chapter 10, Problem 10.158QP

According to information obtained from www.krispykreme.com, a Krispy Kreme original glazed doughnut weighs 52 g and contains 200 Cal and 12 g of fat. (a) Assuming that the fat in the doughnut is metabolized according to the given equation for tristearin

${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}\text{(}\text{s}\text{) + 81}{\text{.5O}}_{\text{2}}\text{(}\text{g}\text{)}\to {\text{57CO}}_{\text{2}}\text{(}\text{g}{\text{) + 55H}}_{\text{2}}\text{O(}\text{l}\text{) }\Delta \text{H}°\text{ = }-37,760\text{ kJ/mol}$

calculate the number of Calories in the reported 12 g of fat in each doughnut, (b) If all the energy contained in a Krispy Kreme doughnut (and just in the fat) were transferred to 6.00 kg of water originally at 25.5°C, what would be the final temperature of the water? (c) When a Krispy Kreme apple fritter weighing 101 g is burned in a bomb calorimeter with C_cal = 95.3 kJ/°C, the measured temperature increase is 16.7°C. Calculate the number of Calories in a Krispy Kreme apple fritter, (d) What would the ΔH° value be for the metabolism of 1 mole of the fat tristearin if the water produced by the reaction were gaseous instead of liquid? [Hint: Use data from Appendix 2 to determine the ΔH° value for the reaction H₂O(l) → H₂O(g).]

Interpretation Introduction

Interpretation:

The number of calories in each doughnut, the final temperature of Water, the number of calories in Krispy Kreme apple fritter and the standard enthalpy of the reaction has to be calculated.

Concept Introduction:

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.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 10.158QP

The number of calories in 12 g fat of each doughnut is $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$.

Record the given info

Weight of doughnut = $\text{52}\text{g}$

Number of calories in $\text{52}\text{g}$ of doughnut = $\text{200}\text{Cal}$

Amount of fat= $\text{12}\text{g}$

Explanation of Solution

The weight of doughnut along with the number of calories and amount of fat is recorded as shown above.

To calculate the number of moles in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$

Molar mass of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$= $\text{891}\text{.45}\text{g}{\text{mol}}^{\text{-1}}$

Number of moles = $\text{12}\text{g×}\frac{\text{1}\text{mol}{\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}}{\text{891}\text{.45}\text{g}}$

= $\text{0}\text{.013}\text{mol}$

Number of moles in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$= $\text{0}\text{.013}\text{mol}$

The number of moles in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is calculated by plugging in the values of mass of fat to the molar mass. The number of moles in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is found to be $\text{0}\text{.013}\text{mol}$.

To calculate the energy of the reaction

The thermochemical equation can be given as,

${\text{C}}_{\text{15}}\text{H}{\text{110}}_{}{\text{O}}_{\text{6(s)}}\text{+81}{\text{.5O}}_{\text{2(g)}}\to {\text{57CO}}_{\text{2(g)}}{\text{+55H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

$\begin{array}{l}\text{(0}\text{.013}\text{mol)}\left[{\text{C}}_{\text{15}}\text{H}{\text{110}}_{}{\text{O}}_{\text{6(s)}}\text{+81}{\text{.5O}}_{\text{2(g)}}\to {\text{57CO}}_{\text{2(g)}}{\text{+55H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\right]\\ \\ \text{and}\\ \\ \text{(0}\text{.013}\text{mol)(ΔH)=(0}\text{.013}\text{mol)(-37,760}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\text{gives}\end{array}$

$\text{0}{\text{.013C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6(s)}}\text{+1}{\text{.06O}}_{\text{2(g)}}\to \text{0}{\text{.740CO}}_{\text{2(g)}}\text{+0}\text{.715}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH=490}\text{.9}\text{kJ}$

$\text{490}\text{.9}\text{kJ}$ of energy is released from the metabolism of $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$

The energy of reaction is calculated from the thermochemical equation and the moles of $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$.  The energy released from the metabolism of $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is found to be $\text{490}\text{.9}\text{kJ}$.

To calculate the number of calories

Number of Calories = $\text{490}\text{.9}\text{kJ×}\frac{{\text{1×10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{×}\frac{\text{1}\text{cal}}{\text{4}\text{.184}\text{J}}\text{×}\frac{\text{1}\text{Cal}}{\text{1000}\text{Cal}}$

= $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$

Number of Calories in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ = $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$

The energy of reaction is calculated from the thermochemical equation and the moles of $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$.  The energy released from the metabolism of $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is found to be $\text{490}\text{.9}\text{kJ}$.

To calculate the number of calories

Number of Calories = $\text{490}\text{.9}\text{kJ×}\frac{{\text{1×10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{×}\frac{\text{1}\text{cal}}{\text{4}\text{.184}\text{J}}\text{×}\frac{\text{1}\text{Cal}}{\text{1000}\text{Cal}}$

= $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$

Number of Calories in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ = $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$

Conclusion

The number of Calories in of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is calculated by plugging in the values of energy released and using the conversion of $\text{kJ}\text{into}\text{Cal}$. The number of Calories in $\text{12}\text{g}$ of ${\text{C}}_{\text{57}}{\text{H}}_{\text{110}}{\text{O}}_{\text{6}}$ is found to be $\text{1}{\text{.2×10}}^{\text{2}}\text{Cal}$.

Interpretation Introduction

Interpretation:

The number of calories in each doughnut, the final temperature of Water, the number of calories in Krispy Kreme apple fritter and the standard enthalpy of the reaction has to be calculated.

Concept Introduction:

Specific heat can be defined as quantity of heat required to raise the temperature of $\text{1}\text{g}$ substance by $\text{1°C}$. The relationship between heat and change in temperature can be expressed by the equation given below.

$\text{q=smΔT}$

Where,

$\text{q=}$ Heat added

s= Specific heat

$\text{m=}$ Mass

$\text{ΔT=}$ Change in temperature.

The unit of specific heat is $\text{J}{\text{g}}^{\text{-1}}\text{.°C}$.

Answer to Problem 10.158QP

The final temperature is found to be $\text{58}\text{.8°C}$

Explanation of Solution

Record the given info

Mass of Water = $\text{6}\text{.00}\text{kg}$

Initial temperature of Water = $\text{25}\text{.5°C}$

The mass of Water along with its initial temperature is recorded as shown above.

To calculate the mass of Water in grams

Mass of Water (in g) = $\text{6}\text{.00}\text{kg×}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)$

= $\text{6000}\text{g}$

Mass of Water (in g) = $\text{6000}\text{g}$

The mass of Water in kilograms is converted into grams by using the conversion factor.

To convert 200 Cal to Joules

= $\text{200}\text{Cal×}\frac{\text{1000Cal}}{\text{1}\text{Cal}}\text{×}\frac{\text{4}\text{.184}\text{}\text{J}}{\text{1}\text{Cal}}\text{=}\text{836800}\text{J}$

200 Cal to Joules = $\text{836800}\text{J}$

200 Cal is converted into Joules using the conversion factor.

To calculate the final temperature

$\text{q=smΔT}$

$\begin{array}{l}\text{836800}\text{J=6000}\text{g×}\left(\frac{\text{4}\text{.184}\text{J}}{\text{g°C}}\right)\text{×}\left({\text{T}}_{\text{f}}\text{-25}\text{.5°C}\right)\\ \\ \text{836800}{\text{J=25104×(T}}_{\text{f}}\text{-25}\text{.5°C)}\\ \\ {\text{T}}_{\text{f}}\text{=58}\text{.8°C}\end{array}$

Final temperature of Water = $\text{58}\text{.8°C}$

The final temperature of Water is calculated by plugging in the values of heat, specific heat of Water and the initial temperature. The final temperature of Water is found to be $\text{58}\text{.8°C}$

Conclusion

The final temperature of Water was calculated by plugging in the values of heat, specific heat of Water and the initial temperature. The final temperature of Water was found to be

$\text{58}\text{.8°C}$

Interpretation Introduction

Interpretation:

The number of calories in each doughnut, the final temperature of Water, the number of calories in Krispy Kreme apple fritter and the standard enthalpy of the reaction has to be calculated.

Concept Introduction:

Heat capacity can be defined as quantity of heat required to raise the temperature of the object by $\text{1°C}$.

The other equation that is used to calculate heat from heat capacity can be given by,

$\text{q=CΔT}$

Where,

$\text{C=}$ Heat capacity of calorimeter

$\text{ΔT=}$ Change in temperature

The unit of heat capacity is $\text{J per°C}$

Answer to Problem 10.158QP

The number of calories in Krispy kreme fitter is 380 Cal.

Explanation of Solution

Record the given info

Heat capacity of calorimeter = $\text{95}\text{.3}\text{kJ}\text{per°C}$

Change in temperature = 16.7^ᵒC

The heat capacity of calorimeter and change in temperature is recorded as shown above.

To calculate the number of Calories in Cal

$\begin{array}{l}{\text{q}}_{\text{reaction}}{\text{=C}}_{\text{Cal}}\text{ΔT}\\ \\ {\text{q}}_{\text{reaction}}\text{=}\left(\text{(95}\text{.3}\text{kJ}\text{per°C)(16}\text{.7°C)}\right)\\ \\ {\text{q}}_{\text{reaction}}\text{=1592}\text{kJ}\\ \\ \text{Convert kJ}\text{into}\text{Cal}\\ \\ \text{1592}\text{kJ×}\frac{{\text{1×10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{×}\frac{\text{1}\text{cal}}{\text{4}\text{.184}\text{J}}\text{×}\frac{\text{1}\text{Cal}}{\text{1000}\text{cal}}\\ \\ \text{=380}\text{Cal}\end{array}$

$Number\text{ of calories = }380\text{ cal}$

The number of Calories is calculated by converting the values of Heat in kJ. The value of Heat in kJ is calculated using the heat capacity of Calorimeter and change in temperature. The number of calories in Cal is found to be $380\text{ cal}$.

Interpretation Introduction

Interpretation:

The number of calories in each doughnut, the final temperature of Water, the number of calories in Krispy Kreme apple fritter and the standard enthalpy of the reaction has to be calculated.

Concept Introduction:

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.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 10.158QP

The change in enthalpy for metabolism for 1 mole of Tristearinis $\text{-35,340}\text{kJ}{\text{mol}}^{\text{-1}}$.

Explanation of Solution

To calculate the enthalpy change :

The chemical equation is,

${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}{\text{O}}_{\left(g\right)}\text{=-241}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}{\text{=ΔH}}_{\text{f}}{\text{°(H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}{\text{)-ΔH}}_{\text{f}}{\text{°(H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=(-241}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)-(-285}\text{.8kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=44}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The change in enthalpy for $\text{55}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}$ is given as,

Change in enthalpy= $\frac{\text{44}\text{kJ}}{\text{mol}}\text{×55}\text{mol=2420}\text{kJ}$

Therefore, the change in enthalpy for one mole is,

Change in enthalpy for one mole= $\text{-37760kJ+2420kJ=-35340}\text{kJ}{\text{mol}}^{\text{-1}}$

Change in enthalpy for one mole= $\text{-35340}\text{kJ}{\text{mol}}^{\text{-1}}$

The change in enthalpy for one mole is calculated by plugging in the sum of values of enthalpy of reaction to the enthalpy of reaction for 55 moles.  The change in enthalpy for one mole is found to be $\text{-35340}\text{kJ}{\text{mol}}^{\text{-1}}$.