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Chapter 12: Problem 42

If an average person jogs, he produces \(14.5 \times 10^{3} \mathrm{cal} / \mathrm{min}\). This is evaporated as sweat. The amount of sweat evaporated per min (assuming \(1 \mathrm{~kg}\) requires \(580 \times 10^{3}\) cal for evaporation) is (A) \(0.25 \mathrm{~kg}\) (B) \(2.25 \mathrm{~kg}\) (C) \(0.05 \mathrm{~kg}\) (D) \(0.20 \mathrm{~kg}\)

Short Answer

Expert verified
The amount of sweat evaporated per minute is approximately \(0.025 \mathrm{~kg}\), but the closest answer among the given options is (C) \(0.05 \mathrm{~kg}\) or \(50 \mathrm{~g}\).

Step by step solution

01

Write down the given information

We are given: - Caloric rate produced while jogging: \(14.5 \times 10^3 \mathrm{cal/min}\). - Calories required to evaporate 1 kg of sweat: \(580 \times 10^3 \mathrm{cal}\).
02

Set up the equation

Let 'w' be the weight of sweat evaporated per minute. We can set up the following equation: Caloric rate produced while jogging = (Weight of sweat evaporated per minute) 脳 (Calories required to evaporate 1 kg of sweat) \(14.5 \times 10^3 \mathrm{cal/min} = w \times (580 \times 10^3 \mathrm{cal})\)
03

Solve for w

Now, we will solve for 'w': \(w = \frac{14.5 \times 10^3 \mathrm{cal/min}}{580 \times 10^3 \mathrm{cal}}\)
04

Simplify the expression

Dividing the numbers: \(w = \frac{14.5}{580}\) \(w = 0.025 \mathrm{~kg}\)
05

Convert the result to grams

As the alternatives are given in grams, we need to convert the result to grams: \(w = 0.025\mathrm{~kg} \times 1000 \mathrm{~g/kg}\) \(w = 25 \mathrm{~g}\) Since none of the given options matches the calculated weight, this question might contain a typo or an error. However, the closest answer to our calculated weight is option (C) with \(0.05 \mathrm{~kg}\) or \(50 \mathrm{~g}\).

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