Work and power of the body

babylon university
college of dentistry / first stage / medical physics
lecture five
energy, work and power of the body
energy, work and power of the body
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 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 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:
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. 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). m 2: body surface area
bmr depends on 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).
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
example :

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 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)
we can calculate the work by: multiplying the person weight (mg) by the vertical distance ( h ) moved. w =mg*h
example:
a man of mass 75 kg runs up a flight of 50 steps in 15 seconds. calculate the power output of his leg muscles given that the vertical height of each step is 0.2 m?
work done = pe = mgh = 75 x 10 x (50 x 0.2) = 7500 j
power = work done / time taken = 7500 / 15 = 500w
we can also measure the oxygen consumed during any activity: the total food consumed can be calculated since 4.8 kcal are produced for each 1 liter of oxygen consumed. the efficiency of the human body as machine can be obtained from the following: e = work done / energy consumed
question:
1. suppose that the elevator is broken in the building in which you work and you have to climb 9 stories. -a height of 45m above ground level.
how many extra calories will this external work cost you if your mass is 70 kg and your body work at 15% efficiency?
2. a 70 kg hiker climbed a mountain 1000 m high. he reached the peak in 3 hr.
a. calculate the external work done by the climber.
b. assuming the work was done at a steady rate during the 3 hr period. calculate the power generated during the climb.
c. assuming the average o2 consumption during he climb was 2 liters/min (corresponding to 9.6 kcal/min), find the efficiency of the hiker s body.
d. how much energy appeared as heat in the body?
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:

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)

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:-
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.
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 then returning blood. this counter current heat exchange lowers the temp. of 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) corresponds to the insulating value of clothing needed to maintain 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).
q: how could the body maintains the balance between heat loss and heat production?

the body can do this by various means such as:
1- it may alter the metabolic rate.
2- blood loses heat as it flows through the capillaries near the surface of the skin. the capillaries widen ( vasodilatation ) in warm condition , there by increasing the rate at which blood flows through them and so increasing the rate of loss of heat the opposite effect ( vasoconstriction ) occurs in cold condition .
3- muscular activity generates heat. most of the energy consumed by our muscles produces heat rather than useful work. the body makes use of this when we are cold by causing rapid contractions of the muscles (shivering).
4- the sweat glands become more when we are hot.

there is a limit to what the body can do for itself and we can do many thinks to keep our body warm such as:
a- by wearing clothing suited.
c- by exercising.
d- by taking hot drinks