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A person is losing thermal energy through the skin at a rate of 120 W when his skin temperature is 30°C. He puts on a sweater, and his skin temperature rises to 33°C. The effective thermal conductivity between his core and the environment changes from 0.22 W/m·K to 0.18 W/m·K. At what rate (J/s) is he now losing thermal energy?

Sagot :

The rate at which the person is now losing thermal energy is 99 J/s.

Thickness of the person's skin

The thickness of the person's skin is calculated as follows;

Q = k(ΔT)h

where;

  • k is thermal conductivity
  • h is thickness of the person's skin
  • ΔT difference in temperature across the skin = 30 °C = 303 K

h = Q/k(ΔT)

h = (120) / (0.22 x 303)

h = 1.8 m

Rate at which the person is now losing thermal energy

Q = k(ΔT)h

where;

  • ΔT is new temperature difference across the skin = 33 °C = 306 K

Q = (0.18)(306)(1.8)

Q = 99 J/s

Thus, the rate at which the person is now losing thermal energy is 99 J/s.

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