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`        Assume that the total surface area of a human body is 1.6 m2 and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant σ is 6.0 × 10-8 W/m2-K4. `
3 years ago

Jitender Pal
365 Points
```										Sol. Q/t = KA(T base – T base 2)/L Rise in Temp. in T base 2 ⇒ KA(T base 1 – T base 2)/Lm base 1s base 1
Fall in Temp in T base 1 ⇒ KA(T base – T base 2)/Lm base 2s base 2 Final Temp. T base 1 – T base 1 - KA(T base 1 – T base 2)/Lm base 1s base 1
Final Temp. T base 2 – T base 2 + KA(T base 1 – T base 2)/Lm base 1s base 1
∆T/dt = T base 1 - KA(T base 1 – T base 2)/Lm base 1s base 1 – T base 2 - KA(T base 1 – T base 2)/Lm base 2s base 2 = (T base 1 – T base 2) – [KA(T base 1 – T base 2)/Lm base 1s base 1 + KA(T base 1 – T base 2)/Lm base 2s base 2]
⇒ dT/dt = - KA(T base 1 – T base 2)/L (1/m base 1s base 1 + 1/m base 2s base 2) ⇒ dT/T base 1 – T base 2 = KA/L(m base 2s base 2 + m base 1s base 1/m base 1s base 1m base 2s base 2)dt
⇒ In∆t = - KA/L (m base 2s base 2 + m base 1s base 1/m base 1s base 1m base 2s base 2)t +C
At time t = 0, T = T base 0, ∆T = ∆T base 0  ⇒ C = In∆T base 0
⇒ In ∆T/∆T base 0 = - KA/L (m base 2s base 2 + m base 1s base 1/m base 1s base 1m base 2s base 2)t ⇒ ∆T/∆T base 0 = e^-KA/L(m base 1s base 1 + m base 2s base 2/m base 1s base 1m base 2s base 2)t
⇒ ∆T = ∆T base 0 e^-KA/L(m base 1s base 1 + m base 2s base 2/m base 1s base 1m base 2s base 2)t = (T base 2 – T base 1) e^-KA/L(m base 1s base 1 + m base 2s base 2/m base 1s base 1m base 2s base 2)t

```
3 years ago
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