a person had a skin temperature of 33degrees celcuis and was in a room where the walls were at 20degrees celcuis.if the emissivity is 1 and the body surface is 1.5meter squad,what is the rate of heat loss due to radiation? alpha=0.000000567W/meter squad k exponent 4
Convert those temperatures to Kelvin. Call them T1(body) and T2(room walls)
The heat loss rate is
Q = (emissivity)*A*(alpha)(T1^4 - T2^4)
where A is the area (1.5 m^2 in this case)
Using this formula implies that the walls have an emissivity of 1 also. The body absorbs a fraction called the absorptivity, which equals the emissivity - assuming no wavelength dependence. That's one of Kirchhoff's radiation laws. There's a lot more to this subject than they are teaching you.
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To calculate the rate of heat loss due to radiation, we can use the Stefan-Boltzmann law, which states that the rate of heat radiation is directly proportional to the fourth power of the temperature difference between the object and its surroundings.
The formula for calculating the rate of heat loss due to radiation is:
Q = ε * α * A * ΔT^4
Where:
Q = Rate of heat loss (in watts)
ε = Emissivity of the object (which is given as 1 in this case)
α = Stefan-Boltzmann constant (which is given as 0.000000567 W/m^2K^4)
A = Surface area of the object (which is given as 1.5 m^2)
ΔT = Temperature difference between the object and its surroundings
Given:
Temperature of the person's skin = 33°C = 33 + 273.15 = 306.15 K
Temperature of the room walls = 20°C = 20 + 273.15 = 293.15 K
ΔT = 306.15 K - 293.15 K = 13 K
Now, we can substitute the given values in the formula to calculate the rate of heat loss due to radiation:
Q = 1 * 0.000000567 * 1.5 * (13)^4
Calculating this expression will give us the rate of heat loss due to radiation in watts.