Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th
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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= 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.