Water of the body

In animals

In humans

Body water

Water molecule

Chemical reactions with water

Water of the body is the water content of an animal body,

contained in the tissues, the blood, the bones, or elsewhere.

Body water makes up a significant fraction of the human

body, both by weight and by volume.

The usual way of adding water to a body is by drinking. In addition, water enters the body with foods, especially those rich in water, such as plants, raw meat, and fish. The amount of this water that is retained in animals is affected by several factors. For instance, animal body water amounts vary with the age of the animal. The older the vertebrate animal, the higher its relative bone mass and the lower its body water content. Water in the animal body performs a number of functions: as a solvent for transportation of nutrients; as a medium for excretion; a means for heat control; as a lubricant for joints; and for shock absorption.

By weight, the average human adult male is approximately 65% water. However, there can be considerable variation in body water percentage based on a number of factors like age, health, weight, and gender. In a large study of adults of all ages and both sexes, the adult human body averaged ~65% water. However, this varied substantially by age, sex, and adiposity (amount of fat in body composition). The figure for water fraction by weight in this sample was found to be 48 ±6% for females and 58 ±8% water for males.

The body water constitutes as much as 75% of the body weight of a newborn infant,

whereas some obese people are as little as 45% water by weight. These figures are

statistical averages, and so like all biostatistics, the estimation of body water

will vary with factors such as type of population, age of people sampled, number

of people sampled, and methodology. So there is not, and cannot be, a figure that is

exactly the same for all people, for this or any other physiological measure.

Water is the chemical substancewith chemical formula H2O:

one molecule of water has twohydrogen atoms covalently bonded

to a single oxygen atom.

oxide I + reducer → oxide II + oxide III

PbO2 + H2 → PbO + H2O

metal oxide + water → hydroxide

K2O + H2O → 2 KOH

base + metalloid oxide → salt + water

KOH + SO2 → K2SO3 + H2O

metal + water → hydroxide + hydrogen

Mg + 2 H2O → Mg(OH)2 +H2↑

nonmetal oxide + water → acid oxygen

SO2 + H2O → H2SO3

metal oxide + acid → salt + water

2 HNO3 + MgO → Mg(NO3)2 + H2O

acid + base → salt + water

HNO3 + NaOH → NaNO3 +H2O

The most common important energy form is heat energy and the conservation of this can be illustrated by considering operations such as heating and drying. In these, enthalpy (total heat) is conserved and as with the mass balances so enthalpy balances can be written round the various items of equipment. or process stages, or round the whole plant, and it is assumed that no appreciable heat is converted to other forms of energy such as work.

In this situation we have to deal with the transmission of heat from the rod into the water. Rod temperature and water - initially different - the heat exchange will have one final temperature (we denote it by tk). This temperature is higher than the initial temperature of water, but less than the initial temperature of the rod.

Here we apply the heat balance - in this particular case, he will have the form:

Q obtained by water = Q rod cast by

Both the heat collected and delivered will be calculated from the formula for heating the heat without change of state:

Q = m · cw · ∆t

Different substances will, however, and the difference in temperature:

Q obtained by water = m water · cw water · (tk - tp water)

Q cast by a rod = m rod · cw steel · (tp rod - tk)

Here are the details:

m water = 6 kg (because a liter of water weighs 1 kg)

m rod = 2 kg

cw steel = 500 J / kg ° C (a read from tables)

water cw = 4200 J / kg ° C (a read from tables)

tk = 70 ° C

Water tp = 20 ° CExplanation

m – masscw – specific heattk – final temperaturetp – initial temperatureQ – heat∆t – increased temperature

We are looking for:

tp rod =?

TransformationsSubstituting the expression for the heat collected and put into the heat balance equation:

m water · cw water · (tk - tp water) = m rod · cw steel · (tp rod - tk)

In the above equation is all information except tp rod.

We divide both sides of the equation by: m rod · cwsteel, then add to both sides of the equation tk. Eventually, you will then searched tp pattern on the rod:

After substituting the numbers we get the final result:

tp bar = 1330 ° C.

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