Question

(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

Answer 1

Answer:

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=$\frac{KA (T_{1}-T_{2})}{x}$   where,

K=Coefficient of thermal conductivity

A=Area of Cross Section

H=Rate of heat loss

x=Thickness of the material

$T_{1}-T_{2}$=Temperature difference between the two surfaces of the body in context

$T_{1}$=37°C ie. Body Temperature

$T_{2}$=20°C ie. Room Temperature

For first case, $T_{1}-T_{2}$=17°C or 17°F

K for fat= 0.21 $Wm^{-1}K^{-1}$

A=2$m^{2}$

x=0.01 m

H=$\frac{0.21 * 2* 17}{0.01}$

H=714 Watts of energy is given out.

2) $Power$×$Time(in seconds) = Energy$

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 $T_{2}=0$ °C

E=$\frac{614.34*37}{17}$=1337.09 KCal

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