Chemistry, 07.05.2020 05:12 marlesly87

The fuel value of hamburger is approximately 3.3 kcal/g. If a man eats 0.75 pounds of hamburger for lunch and none of the energy is stored in his body, estimate the amount of water that would have to be lost in perspiration to keep his body temperature constant. The heat of vaporization of water may be taken as 2.41 kJ/g.

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Answer from: Savageman4654

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/gamount of hamburger consumed, m = 0.75 lbheat of vaporization of water = 2.41 kJ/g

The energy stored in the hamburger is calculated as follows;

Q = Hm

454 g = 1 lb

4.186 kJ = 1 kcal

$Q = (3.3 \ \frac{kcal}{g} ) \times \frac{4.186 \ kJ}{1 \ k cal} \times (0.75 \ lb) \times \frac{454 \ g}{1 \ lb} \\\\Q = 4,703.6 \ kJ$

The mass of water lost to perspiration to maintain the temperature is calculated as follows;

$Q = mc\Delta t \ + \ mL_v\\\\Q = mc (t_2 - t_1) \ + \ mL_v$

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

$Q = m[c(t_2 -t_1) \ + \ Lv]\\\\m = \frac{Q}{c(t_2 -t_1) \ + \ Lv} \\\\m = \frac{4,703.6 \times 10^3 }{4.186 (100-37) \ + \ 2.41 \times 10^3} \\\\m = 1,759.2 \ g\\\\m = 1.76 \ kg$

Thus, the mass of water lost to perspiration to maintain the temperature is 1.76 kg.

Learn more about heat capacity here: link

Answer from: gracekerleyy

$m_{w} = 1755.323\,g$

Explanation:

The energy stored in the hamburger is:

$Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)$

$Q = 4693.239\,kJ$

The amount of water perspired is derived from the following formula:

$Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})$

$m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}$

$m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }$

$m_{w} = 1755.323\,g$

Answer from: Quest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

Answer from: Quest

What are the choices?

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