If you drink a 20-ounce bottle of water (590 g) that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?

Note:
- Your body temperature is 36.6 °C.
- Assume the thermal properties of sweat are the same as for water.
- [tex]C_s[/tex], liquid water = 4.184 J/g °C
- [tex]C_s[/tex], steam = 1.84 J/g °C
- [tex]C_s[/tex], ice = 2.09 J/g °C
- [tex]\Delta H_{\text{vap}}[/tex] = 40.67 kJ/mol at 36.6 °C
- [tex]\Delta H_{\text{fus}}[/tex] = 6.01 kJ/mol

Options:
A. 1413 kJ
B. 81 kJ
C. 1150 kJ

To convert all the water from the 20-ounce bottle (590g) into sweat and then vapor, a total of 81 kJ + 1330.3291 kJ = 1411.3291 kJ of heat is needed.

To calculate the amount of heat needed to convert all the water into sweat and then vapor, we need to consider the different stages of the process. First, we need to raise the temperature of the water from 3.8 °C to 36.6 °C. Then, we need to convert the liquid water into sweat, and finally, we need to convert the sweat into vapor.

1. Raising the temperature:
The specific heat capacity of liquid water (Cs) is given as 4.184 J/g °C. To raise the temperature of the water from 3.8 °C to 36.6 °C, we need to calculate the heat required using the formula:
Heat = mass × specific heat capacity × change in temperature.

The mass of the water is given as 590g, the specific heat capacity of liquid water is 4.184 J/g °C, and the change in temperature is 36.6 °C - 3.8 °C = 32.8 °C.

So, the heat required to raise the temperature of the water is:
Heat = 590g × 4.184 J/g °C × 32.8 °C

= 81857.312 J.

2. Conversion of liquid water to sweat:
Since the thermal properties of sweat are assumed to be the same as for water, the heat required to convert the liquid water into sweat is equal to the heat required to raise the temperature.

So, the heat required to convert water into sweat is also 81857.312 J.

3. Conversion of sweat into vapor:
To convert the sweat into vapor, we need to consider the heat of vaporization (AHvap). The heat of vaporization is given as 40.67 kJ/mol at 36.6 °C.

First, we need to calculate the number of moles of water in 590g using the molar mass of water (18.015 g/mol):
Number of moles = mass / molar mass

= 590g / 18.015 g/mol =

32.73 mol.

The heat required to convert sweat into vapor is calculated using the formula:
Heat = number of moles × heat of vaporization.

Heat = 32.73 mol × 40.67 kJ/mol

= 1330.3291 kJ.

To convert all the water from the 20-ounce bottle (590g) into sweat and then vapor, a total of 81 kJ + 1330.3291 kJ = 1411.3291 kJ of heat is needed.

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