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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.
Learn more about thermal energy here: https://brainly.com/question/7541718
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