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

The rate of heat loss for a lightly dressed person sitting in a comfortable room is approximately 68 Watts. The number of Calories burned in an hour is 0.024 kcal for someone at rest. When sitting outside in freezing temperatures, the body may burn an additional 50-100 Calories per hour.

Step-by-step explanation:

The rate of heat loss in watts for someone who is lightly dressed and sitting in a comfortable room with a temperature of 20°C can be estimated using the formula Q = kAΔT/d, where Q is the rate of heat loss, k is the thermal conductivity of fat (0.2 W/mK), A is the surface area (2 m²), ΔT is the temperature difference between the body and the room (37 - 20 = 17°C), and d is the thickness of the fat layer (1 cm = 0.01 m).

Using this formula, the rate of heat loss can be calculated as follows:

Q = (0.2 W/mK)(2 m²)(17°C)/(0.01 m)

Q = 68 W

Therefore, the rate of heat loss for someone lightly dressed in a comfortable room is approximately 68 Watts.

To calculate the number of Calories (kilocals) burned in an hour, we can use the formula:

Calories = Power (Watts) x Time (hours) x Conversion factor (1 kcal = 4184 J)

If the person is at rest, the amount of heat dissipated is equivalent to the metabolic rate, which is about 100 W. Therefore, the number of Calories burned in an hour is:

Calories = 100 W x 1 hour x (1 kcal/4184 J)

Calories = 0.024 kcal

If the person is sitting outside when it is freezing, the body will burn more Calories to maintain its normal temperature. The exact amount of Calories burned will depend on factors such as the temperature and wind speed. On average, a person may burn an additional 50-100 Calories per hour when exposed to cold temperatures.

Answer 2

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.

Step-by-step explanation:

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

H=
`(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=
`(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=
`(614.34*37)/(17)`=1337.09 KCal

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