Answer :
Answer:
[tex]m_{w} = 1755.323\,g[/tex]
Explanation:
The energy stored in the hamburger is:
[tex]Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)[/tex]
[tex]Q = 4693.239\,kJ[/tex]
The amount of water perspired is derived from the following formula:
[tex]Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})[/tex]
[tex]m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}[/tex]
[tex]m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }[/tex]
[tex]m_{w} = 1755.323\,g[/tex]
The mass of water lost to perspiration to maintain the temperature is 1.76 kg.
The given parameters;
- fuel value of the hamburger, H = 3.3 kcal/g
- amount of hamburger consumed, m = 0.75 lb
- heat of vaporization of water = 2.41 kJ/g
The energy stored in the hamburger is calculated as follows;
[tex]Q = Hm[/tex]
454 g = 1 lb
4.186 kJ = 1 kcal
[tex]Q = (3.3 \ \frac{kcal}{g} ) \times \frac{4.186 \ kJ}{1 \ k cal} \times (0.75 \ lb) \times \frac{454 \ g}{1 \ lb} \\\\Q = 4,703.6 \ kJ[/tex]
The mass of water lost to perspiration to maintain the temperature is calculated as follows;
[tex]Q = mc\Delta t \ + \ mL_v\\\\Q = mc (t_2 - t_1) \ + \ mL_v[/tex]
where;
c is the specific heat capacity of water = 4.186 J/g⁰C
t₂ is the boiling point of water = 100⁰C
t₁ is normal body temperature = 37⁰C
[tex]Q = m[c(t_2 -t_1) \ + \ Lv]\\\\m = \frac{Q}{c(t_2 -t_1) \ + \ Lv} \\\\m = \frac{4,703.6 \times 10^3 }{4.186 (100-37) \ + \ 2.41 \times 10^3} \\\\m = 1,759.2 \ g\\\\m = 1.76 \ kg[/tex]
Thus, the mass of water lost to perspiration to maintain the temperature is 1.76 kg.
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