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Chapter 14: Q26P (page 390)

(II)During exercise, a person may give off 185 kcal of heat in 25 min by evaporation of water (at 20°C) from the skin. How much water has been lost? [Hint: See page 399.]

Short Answer

Expert verified

The amount of water lost is \(0.31\;{\rm{kg}}\).

Step by step solution

01

Given data

The amount of heat is\(Q = 185\;{\rm{kcal}}\).

The time is\(t = 25\;{\rm{min}}\).

The temperature is \(T = 20\circ {\rm{C}}\).

02

Understanding the heat of vaporization

In this problem, use the heat of vaporization at room temperature, whose value is 585 kcal/kg. Consider that the heat from the person is only utilized to evaporate the water.

03

Calculation of the water that has been lost

The relation to find the amount of water lost is given by:

\(Q = mL\)

Here,\(L\)is the heat of vaporization and\(m\)is the mass of water lost.

On plugging the values in the above relation, you get:

\(\begin{array}{c}185\;{\rm{kcal}} = m\left( {585\;{\rm{kcal/kg}}} \right)\\m = 0.31\;{\rm{kg}}\end{array}\)

Thus, \(m = 0.31\;{\rm{kg}}\) is the required amount of water.

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Most popular questions from this chapter

(II)The heat capacity, C, of an object is defined as the amount of heat needed to raise its temperature by 1 C°. Thus, to raise the temperature by \(\Delta T\) requires heat Q given by .

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(c) Of 45 kg of water?

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