Question

2 An average human has about 67 % water by mass. A 75.0 kg person suffering from hypothermia has a body water temperature of 33.5°C and needs to raise this to the normal body temperature of 37.0°C. a) How many grams of sucrose, C12H22O11(s), must be metabolized (use this as a simple combustion reaction with gaseous oxygen, producing liquid water and gaseous carbon dioxide)? 1 4 a) 4.74y lb g sucrose b) What volume of air (L) with a l atm total pressure and a partial pressure of oxygen of 0.2 atm would be needed for this? [Assume the air is 298 K.] b) 1.03 x 10 L air

Answer 1

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SOLUTION:

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PART (a):

Given that, Mass of person, M = 75.0 kg.

Mass percentage of water in human body = 67 %

Therefore, Mass of water in the body is given by,

$m=67\%\times75\;kg=0.5=67\times75\;kg$

m=50.25 kg

We know that,

Heat capacity of water, $C=4186\;J/kg.^{o}C$

Temperature difference required, $\Delta T=37^{o}C-33.5^{o}C=3.5^{o}C$

Therefore, Heat required to raise the temperature of water is given by,

$H=mC\Delta T=50.25\;kg\times \left (4186\;J/kg.^{o}C \right )\times3.5^{o}C$

$\therefore H=736212.75\;J$

We know that, 1 J = 10^-3 kJ.

$\therefore H=736212.75\times10^{-3}\;kJ$

$\boldsymbol{\therefore H=736.21275\;kJ}$

Balanced Combustion reaction of Sucrose is given by,

$C_{12}H_{22}O_{11}_{(s)}+12O_{2}_{(g)}\rightarrow 12CO_{2}_{(g)}+11H_{2}O_{(g)}$

Enthalpy/energy released by this reaction is: $\Delta H=-5645\;kJ/mol$

(negative sign represents energy is released).

This energy will be used to raise the temperature of water.

Therefore, No.of moles of Sucrose required to raise the temperature:

$n=\frac{H}{\Delta H}=\frac{736.21275\;kJ}{5645\;kJ/mol}$

$\therefore n=0.13042\;mol$

We have, molar mass of sucrose = 342.2965 g/mol

Therefore, mass of sucrose needed is given by:

$m= n\times molar\;mass=0.13042\;mol\times 342.2965\; g/mol$

$\boldsymbol{\therefore m= 44.642\;g\;of\;sucrose}$

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PART (b):

From the above balanced Combustion reaction of Sucrose, we have, 1 mol of sucrose reacts with 12 moles of oxygen.

$1\;mol\;of\;sucrose\rightarrow 12\;moles\;of\;oxygen$

Therefore, 0.13042 mol of sucrose will require:

$n=0.13042\;mol\times 12=\boldsymbol{1.565\;mol\;of\;O_{2}}$

We have,

Pressure of O₂= 0.2 atm.

Temperature of O₂= 298 K.

Gas constant, R = 0.0821 L.atm/(mol.K).

Therefore, According to ideal gas equation, Volume of O₂ is given by,

$PV=nRT$

$\therefore V=\frac{nRT}{P}=\frac{1.565\;mol\times (0.0821\;L.atm/mol.K)\times298\;K}{0.2\;atm}$

$\boldsymbol{\therefore V=191.45\;L\;of\;O_{2}}$​​​​​​​

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