How much water would have to evaporate from the skin per minute to make away all the heat generated by the basal metabolism (60 kcal/h) of a 65 kg person?

Answer

To solve the problem of how much water would need to evaporate from the skin per minute to dissipate the heat generated by the basal metabolism of a 65 kg person, we will follow these steps:

### Step 1: Convert the heat generated per hour to heat generated per minute
The heat generated by basal metabolism is given as 60 kcal/h. To find the heat generated per minute, we can convert hours to minutes.

\[ \text{Heat generated per minute} = \frac{60 \text{ kcal}}{60 \text{ min}} = 1 \text{ kcal/min} \]

### Step 2: Convert kcal to calories
Since 1 kcal is equivalent to 1000 calories, we convert the heat generated per minute into calories.

\[ \text{Heat generated per minute} = 1 \text{ kcal/min} \times 1000 \text{ cal/kcal} = 1000 \text{ cal/min} \]

### Step 3: Use the latent heat of vaporization
The latent heat of vaporization of water is approximately 540 kcal/kg (or 540 cal/g). This is the amount of heat required to evaporate 1 gram of water.

### Step 4: Calculate the amount of water needed to evaporate
To find out how much water must evaporate to absorb 1000 calories, we can use the formula:

\[ \text{Amount of water (g)} = \frac{\text{Heat required (cal)}}{\text{Latent heat of vaporization (cal/g)}} \]

Substituting the values we have:

\[ \text{Amount of water (g)} = \frac{1000 \text{ cal}}{540 \text{ cal/g}} \approx 1.85 \text{ g} \]

### Step 5: Round off the answer
We can round off the answer to two decimal places:

\[ \text{Amount of water (g)} \approx 1.85 \text{ g} \]

### Final Answer
Approximately 1.85 grams of water would need to evaporate from the skin per minute to dissipate the heat generated by the basal metabolism of a 65 kg person.

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