We’re being asked to calculate the **amount of heat required to convert the 20 oz of water to vapor**

Converting 20 oz to grams (1 oz = 20.35 g)

$20\overline{)\mathrm{oz}}\left(\frac{28.35\mathrm{g}}{1\overline{)\mathrm{oz}}}\right)=567\mathrm{g}$

1. **q _{1}** which is the heat in raising the temperature of 567 g of water from

2. **q _{2}** which is the heat in raising the temperature of 567 g of water from

3. **q _{3}** which is the heat in evaporating 567 g of water at

We need to solve for each heat individually then add them together to get the final answer.

**For q _{1}:** The temperature changes from 3.8 ˚C to 36.8 ˚C. Because of this, the equation we’ll use is:

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}$

where **m** = mass (in g), **c _{liq}** = specific heat of water (in J/g • ˚C), and

We’re given the following values:

m = 567 **g** c_{liq} = **4.18 J/g • ˚C **ΔT = 36.8 ˚C – 3.8 ˚C = **32.8 ˚C**

__Calculating for __**q _{1}**

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}(567\overline{)g})(4.18\frac{J}{\overline{)g\xb0C}})(32.8\overline{)\xb0C})\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{77}\mathbf{x}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{J}$

**For q _{2}:** The temperature changes from 36.8 ˚C to 100 ˚C. Because of this, the equation we’ll use is:

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}$

where **m** = mass (in g), **c _{liq}** = specific heat of water (in J/g • ˚C), and

We’re given the following values:

m = 567 **g** c_{liq} = **4.18 J/g • ˚C **ΔT = 100˚C - 36.8 ˚C = **63.2 ˚C**

__Calculating for __**q _{2}**

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{567}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\frac{\mathbf{J}}{\overline{)\mathbf{g}\mathbf{\xb0}\mathbf{C}}}\mathbf{)}\mathbf{(}\mathbf{63}\mathbf{.}\mathbf{2}\overline{)\mathbf{\xb0}\mathbf{C}}\mathbf{)}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{x}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{J}$

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water 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. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Heating and Cooling Curves concept. You can view video lessons to learn Heating and Cooling Curves. Or if you need more Heating and Cooling Curves practice, you can also practice Heating and Cooling Curves practice problems.

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