Evaporating 1 gram of water cools 539 grams of a person's body by 1 Celsius degree. How much water, in grams, evaporates to cool an 80-kilogram person by 1 Celsius degree? Round your answer to three significant digits. Round to the nearest gram.

To solve the problem, we can use the relationship given in the problem: evaporating 1 gram of water cools 539 grams of a person's body by 1 Celsius degree. We need to find out how much water needs to evaporate to cool an 80-kilogram (80,000 grams) person by 1 Celsius degree.

First, let's set up the proportion based on the given relationship:

\[
\frac{1 \text{ gram of water}}{539 \text{ grams of body}} = \frac{x \text{ grams of water}}{80,000 \text{ grams of body}}
\]

We can solve for \(x\) by cross-multiplying:

\[
1 \times 80,000 = 539 \times x
\]

\[
80,000 = 539x
\]

Solving for \(x\), we divide both sides by 539:

\[
x = \frac{80,000}{539}
\]

\[
x \approx 148.425
\]

Rounding to the nearest gram, the amount of water that needs to evaporate to cool an 80-kilogram person by 1 Celsius degree is:

\[
x \approx 148 \text{ grams}
\]

So, the answer is 148 grams.

Would you like more details or have any questions about this solution? 

Here are 5 questions you might consider asking next:
1. How do you derive the formula for the proportion in this type of problem?
2. Can you explain the concept of specific heat capacity and its role in cooling?
3. How does the process of evaporation contribute to cooling in terms of energy transfer?
4. What other factors might affect the rate of evaporation and cooling in real-life scenarios?
5. Can you provide examples of similar problems involving cooling and heating?

**Tip:** When working with proportions, setting up the ratio correctly is crucial. Always double-check the units to ensure consistency across your calculations.

Learn how to calculate the amount of water needed to cool an 80-kilogram person by 1 Celsius degree through evaporation. Detailed step-by-step solution provided, rounding to three significant digits. Suitable for students in Grades 9-12.

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