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pychu [463]
3 years ago
6

A human subject with mass 96 kg, body specific heat 3500 Jkg^-1K^-1 skin temperature 34.5 degrees Celsius, surface area 1.5m^2 a

nd emissivity 0.7 enters a room whose walls are at temperature 21 degrees Celsius. At what rate does body temperature decrease? Answer in degrees per hour, as a positive number
Physics
1 answer:
AlekseyPX3 years ago
7 0

rate of heat radiation by the body is given by

\frac{dQ}{dt} = \sigma e A(T^4 - T_s^4)

here we know that

\sigma = 5.67 \times 10^{-8}

e = 0.7

A = 1.5 m^2

T = 34.5 + 273 = 307.5 k

T_s = 21 + 273 = 294 k

now from above formula rate of heat dissipation

\frac{dQ}{dt} = (5.67 \times 10^{-8}}(0.7)(1.5)(307.5^4 - 294^4)

\frac{dQ}{dt} = 87.5

now we know that

Q = ms \Delta T

from above equation

\frac{dQ}{dt} = ms\frac{dT}{dt}

now we have

m s \frac{dT}{dt} = 87.5

here we have

m = 96 kg

s = 3500

96 \times 3500 \times \frac{dT}{dt} = 87.5

\frac{dT}{dt} = (2.6 \times 10^{-4}) \: ^0C/s

\frac{dT}{dt} = 0.94 ^0 C/hour

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2 years ago
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Hope that was helpful!!

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