Energy work and power of the body

energy, work and power of the body
أستاذ المادة: المدرس ناهدة حمود الج ا رح
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all body activities including thinking, doing work, or keeping the body temp. constant involve energy changes,
for example under resting(basal)conditions the skeletal muscles and the heart using 25% of the body s energy
,another 19%is being used by the brain,10%is being used by the kidneys, and 27% is being used by the liver and the
spleen. a small percent of about 5% of food energy being excreted in feces and urine.
extra food energy will be stored mainly as fat. external heat energy from environment can help maintain the
body temp. , but it has no use in body function.
conservation of energy
change in the stored energy (i.e. food energy, body fat and the body heat)
=heat lost from the body + work done
assumes that no food or drink is taken and no feces or urine is excreted during the interval of time considered.
?this is similar to the first law of thermodynamic:-
?q= ?u + ?w
?where ?q is the change of quantity of heat of the system.
?u is the change in the internal or stored energy.
?w is the work done.
this can be written as
?u= ?q - ?w
a body doing no work (?w=0) and at constant temp. continues to lose heat to its surroundings, and ?q is negative.
therefore, ?u is
also negative, indicating a decrease in stored energy.
the rate of change of their variables is just taken per unit time
( by dividing on ?t) .
?u / ?t = ?q/?t - ?w/?t
the body s basic source of energy is the food energy it must be chemically changed by the body to make
molecules that can combine with oxygen in the body s cells.
energy change in the body
the units are joule or calorie 1cal= 4.184j or 1kcal=4184j
the power is defined as energy or work per unit time =j/s=watt.
in the oxidation process within the body, heat is produced as energy of metabolism. the rate of oxidation is
called metabolic rate.
for example the oxidation of one mole of glucose can be shown as:
c6h12o6 + 6o2 6h2o + 6co2 + 686kcal
1 mole 6 mole 6 mole 6 mole
180g 192g 108g 264g
co2 and o2 are gases ( 1 mole of a gas at normal temp. and pressure has a volume 22.4 liters)
from the above equation we can calculate useful quantities for glucose metabolism:
?kcal of energy released/g of fuel (glucose) =686/180=3.8
?kcal of energy released/l of o2 used=686/ (22.4×6) =5.1
?liters of o2 used/g of fuel glucose = (22.4×6)/180=0.75
?liters of co2 produced /g of fuel glucose= (22.4×6)/180=0.75
so the ratio of moles of co2 produced to moles of o2 used, called the (respiratory quotient) r=1
no. of moles of co2/no. of moles of o2 =1
similar calculation can be done for fats, proteins, and other carbohydrates.
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by measuring the energy released per liter of o2 we can get a good estimation of the energy released. table 5.1
shows the caloric values for different types of foods and fuels. it gives the maximum values expected because not all
food energy is available, as part of it is lost in incomplete combustion (not metabolized).
when the body is completely at rest, it will have the lowest rate of energy consumption this is called the basal
metabolic rate (bmr), which is the amount of energy needed to perform minimal body functions (such as breathing
and pumping the blood through the arteries) under resting conditions, and for typical person 92 kcal/hr
? 107w or about 1 met (met is 50 kcal/m2hr).
m2: body surface area
bmr depends on sex, age, height, and weight it depends primarily on thyroid function, overactive thyroid gives higher
bmr.
since the energy used for basal metabolism becomes heat which is mainly dissipated from the skin, so the basal rate is
related to the surface area or to the mass of the body. in figure 5.1 the graph represents the change between bmr (kcal/day)
and the mass of different animals, the slope of the graph indicates that the bmr is proportional to mass.
when the animals gets larger the bmr increases faster than their increases in surface area but bmr increases
even more faster with their mass(volume).
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the bmr depends to large extent on the body temp., for an increase of 1°c it will change by 10% in the metabolic
rate, so for 3°c the change will be 30% greater than normal. similarly ,if the body temp. dropings 3°c below normal, the
metabolic rate decreases by about 30%. for this reason hibernating animals at low body temp. will reduce
the metabolic rate very much. a man who is taking food energy equivalent to his bmr plus his other
physical activities will keep on constant weight.
less food will cause weight lose and for longer time cause starvation.
excess food of body needs will cause food storage and increase in weight.
bmr is sometimes determined from oxygen consumption when resting, we can also estimate the food energy used in
various physical activities by measuring the oxygen consumption, table (5.2) shows some typical values for
various activities.
example: suppose you wish to lose 4.54kg either through physical activity or by dieting.
1-how long would you have to work at an activity of 15kcal/min to
lose 4.54kg of fat?
from table 5.1 maximum of 9.3kcal/g of fat, if you worked for t minutes, then
t(15kcal/min)=(4.54x103g)(9.3kcal/g)=4.2x104kcal
t=2810 min?47hr
2- it is much easier to lose weight by reducing your food intake. if you normally use 2500kcal/day, how long must you
diet at 2000kcal/day to lose 4.54kg of fat?
t= (energy of 4.54kg fat/energy deficit per day) 4.2x104kcal / 5x102kcal/day ? 84 days
table 5.3 gives oxygen consumption for various organs, some organs use rather large amount of energy ,the kidney
uses more energy per kilogram than the heart.
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work and power
chemical energy stored in the body is converted into external
mechanical work as well as into life-preserving functions.
mechanical work is usually defined by ?w=f. ?x where f is the force on the same line of displacement x, or it can
be also written as:
(?w= f ?x cos ?) where ? is the angle between f and the direction
of movement, the power is work per unit time.
p= ?w/?t = f ?x / ?t = f v where v is the velocity
when the force is perpendicular to the displacement work will be zero, such as walking body, his weight is
perpendicular to distance of movement but practically it will not be zero because the uses energy against friction and
other movement of his body, but in the case of climbing person for distance (h) the weight is on the same line of
displacement then the work = mgh, the efficiency of human body is
e = work done/ energy consumed
efficiency is usually lowest at low power but can increase to 20% for trained individuals in activities such as
cycling and rowing.
the maximum work capacity of the body is variable, for short period of time the body can perform at very high
power levels,(like running very fast but it is more limited for longer periods).it is found that long term power is
proportional to the maximum rate of oxygen consumption in the working muscles.
for healthy man this consumption is 50ml/kg m of body weight each minute. the body can supply an instantaneous
energy for short term power needs, this can be done by splitting energy rich-phosphates and glycogen leaving an
oxygen deficit in the body. this process can only last about a minute and is called anaerobic(without oxygen) .for
longer term work requires oxygen aerobic. fig (5.3)
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heat loses from the body
(homeothermic)(warm-blooded) such as birds and mammals, (poikilothermic)(cold-blooded) such as frog and
snake, will have a higher body temp. on a hot day than mammals, birds and mammals both have mechanisms to
keep their body temp. constant despite fluctuations in the environmental temp. constant body temp. permit
metabolic processes to produce at constant rates and these animals to remain active even in cold climates.
the normal human temp. is 37°c which is obtained from taking the temp. of large # of people. for a single
individual the body temp. may vary about ?0.5°c.the rectal temp. is about 0.5°c higher than the oral temp.
the temp. of the body depends on the:
1-time of the day (lower in the morning)
2- environment temp.
3-the amount of clothing
4-health of the person
5- on his recent physical activity.
for example rectal temp. after hard exercise may be as high as 40°c ,the body losses heat mainly by radiation,
convection, and evaporation, all these processes can take place in the skin. the evaporation of perspiration from the
skin can cool down the skin by absorbing the latent heat of evaporation from it.
evaporation takes place also in breathing causing cooling effect. if the air is cold it will also cool down the
body. eating and drinking cold or hot food can also decrease or increase the body temp.
the body temp. is kept constant for this reason the hypothalamus in the brain can control the body temp.
(thermostate like).after heavy exercise the body is heated the hypothalamus initiate the sweating and vasodilation is
the first causes heat loss by evaporation and the second increasing the blood supply to the skin for more loss of heat.
on the other hand if the environment temp. dropings the thermo receptors on the skin signals to the hypothalamus
which in turn induce shivering to increase the body temp. the production of heat in the body for 2400
kcal/day (assumeing no change in body weight)=1.7kcal/min=120j/sec =120w.
so the body must lose the same amount of heat to stay at constant temp. .the heat loses depends on many
factors:
1- the temp. of the surroundings 2-humidity 3-motion of the air 4-the physical activity of the body 5-the
amount of the body exposed 6-the amount of the insulation of the body (like clothes and fat)
transfer of heat by radiation
all objects regardless on their temp. emit electromagnetic radiation, the amount of energy emitted by the body is
proportional to the absolute temp. raised to the fourth power. the body also receives radiant energy from surrounding
objects. the amount of heat difference between the energy radiated by the body and the energy absorbed from the
surrounding can be calculated from the equation:
hr = kr ar e (ts-tw)
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where
(hr) is the rate of heat energy loss or gain
(kr) is a constant depends upon various physical parameters and it s about =5kcal/m2 hr c° for man
(ar) effective body surface area emitting radiation
e is the emissivity of the surface which is nearly=1,independent on the color of the skin indicating that the skin at
this wavelength is almost a perfect emitter and absorber of radiation.
(ts) is the skin temp. in c°
(tw) is the temp of the surrounding walls
? heat losses by radiation occur even the temp. differences is not high.
example: for a nude person have a skin temp. 34°c in a room of walls temp. 25°c and his body area 1.2m2 will lose 54
kcal/hr which is 54% of the total losses. most of the remaining heat will be by convection.
transfer of heat by convection
heat losses by convection (hc)
hc = kc ac (ts-ta)
where:
hc is the amount of heat gained or lost be convection
ac is the effective surface area ts is the skin temp.
ta is the environment temp. or air temp.
kc is a constant that depends on the movement of the air, for a resting body and no apparent wind kc is about 2.3kcal/
m2 hr °c. when the air is moving kc increases according to the equation:-
kc = 10.45 – v + 10? v where v is the wind speed in m/sec
this equation is valid for speeds between 2.23m/sec (5mph) and 20m/sec (45mph) (1 mile=1.6 km).
the equivalent temp. due to moving air is called the wind chill factor and is determined by the actual temp. and
wind speed. for example for a windy day speed 10 m/sec an-20°c has the same cooling effect on the body as -40°c
on a calm day. table (5.5)
transfer of heat by evaporation
under normal temp. conditions and in the absence of hard work or exercise, heat loss mainly by radiation and
convection, losses by evaporation become of less importance. under extreme conditions of heat and exercise, a man
may sweat more than 1 litter of liquid per hour. since each gram of water that evaporate carries with it the heat of
vaporization of 580 calories, the evaporation of i liter carries with it 580kcal. there is some heat losses by
perspiration even if the body does not feel sweaty, it amount to about 7kcal/hr, equivalent to 7% of the body losses.
a similar loss of heat is due to the evaporation of moisture in the lungs, an additional amount of water will be
evaporated during expiration. this will cool the body the same as the evaporation from the skin, also when we
inspire cold air inside the lungs which also cool down the body. under typical conditions the total respiratory heat
losses is about 14% of the body s heat loss.
under extreme condition of heat and exercise the sweat evaporation is very important, a man may sweat more than
1 lit/hr, this is if all sweat is evaporated, the un evaporated part (running down)does not contribute with cooling.
counter current heat exchange
since the radiation of heat from the body and the transfer of heat to the air depend upon the skin temp., any factors
that affect the skin temp. also affect the heat loss. the body has the ability to select the path returning blood from the
hands and feet. in cold weather blood is returned to the heart through internal veins that are in contract to the arteries
carrying blood to the extremities (hands and feet).in this way some of the heat from the blood going to the extremities
is used to heat thee returning blood. this counter current heat exchange lowers the temp. of the extremities and reduces
heat loss from the body to the environment.
in warm weather the returning venous blood runs near the skin surface raising the skin temp. and thus increasing
the heat loss from the body.
most of the previous study involved heat losses from a nude person, if we consider the clothes, the calculation
become more complicated, for this reason another unit of clothing is the (clo) is being introduced. one (clo)
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corresponds to the insulating value of clothing needed to maintain a subject sitting at rest in comfort in a room at 21°c
and air movement of 0.1m/sec and humidity of less than 50%.
one (clo) is equivalent to lightweight suit an individual in the arctic needs clothing of insulation of 4 clos.
(a fox fur has an insulating value of 6 clos).