(1) a typical person has a surface area of about 2m^2 and a layer of fat about 1cm thick (it can vary from 0.5 cm to 1.5 cm) . estimate the rate of heat loss in watts for someone who is lightly dressed sitting in a comfortable room with a temperature of 20 c.(2) how many calories (kilocals) does the person burn in an hour? (3) how many calories (kilocals) does a person burn in a hour if they are sitting outside when it is freezing

1)H=714 Watts of energy is given out.

2)614.34 KCal energy is burnt in an hour in the room

3)In freezing conditions, 1337.09 KCal is spent in an hour.

Explanation:

The formula for thermal conductivity states that the rate of heat loss H is as follows:

H=   where,

K=Coefficient of thermal conductivity

A=Area of Cross Section

H=Rate of heat loss

x=Thickness of the material

=Temperature difference between the two surfaces of the body in context

=37°C ie. Body Temperature

=20°C ie. Room Temperature

For first case, =17°C or 17°F

K for fat= 0.21

A=2

x=0.01 m

H=

H=714 Watts of energy is given out.

2) ×

Energy= 714 × 60 × 60=2570400 J

2570400 J=614.3403442 KCal

Therefore, 614.3403442 KCal energy is burnt in an hour.

3) Only the temperature difference becomes 37°C from 17°C , rest everything remains the same, therefore the energy will also vary due to the temperature factor and more energy will be spent as in freezing climate °C

E==1337.09 KCal

In freezing conditions, 1337.09 KCal is spent in an hour.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be