Question

During a skin burn from an oven, approximate the skin and tissue layer to be infinitely thick compared to the damaged layer. The temperature throughout the entire skin and tissue layer is uniform at 33ºC before contact with the oven, and the surface layer of skin increases to the temperature of the oven, 200ºC, instantaneously upon contact. Find the depth of the damaged layer of skin after 2 seconds of exposure to the oven temperature. Consider skin as being damaged when it reaches 62ºC. The thermal diffusivity of skin is 2.5 x 10-7 m2/s.

Answer 1

The depth of the damage is given as
`x = 1.357*10^(-3)m`

Step-by-step explanation:

From Fick's second law

`(C_(x)-C_(0))/(C_(s)-C_(0)) = 1 -erf [(x)/(2√(Dt) ) ] ...(1)`

Where

`C_x` is the temperature when the skin is damage

`C_0` is the initial temperature of the skin

`Cs` is the temperature of the oven

erf is a mathematical function known as error function

D is the thermal diffusivity

x is the depth of the damage

From the question we are given that

`C_x` = 62°C

`C_0` = 33°C

D =
`2.5 * 10^(-7) m^2/s.`

Substituting values into equation one

`(62-33)/(200-33) = 1 -erf [\frac{x}{2\sqrt{2.5*10^(-7)} *2} ]`

`erf[(x)/(1.414*10^(-3)) ] = 0.826\\`

Let assume that
`z =(x)/(1.414*10^(-3)) ... (2)`

Hence

`erf(z) = 0.826`

Looking at the table of error function shown on the first uploaded image

We can use interpolation to obtain the value of z in this manner

`(z-0.95)/(0.8260-0.8209) = 1`

z = 0.96

Substituting the value of z into equation 2

`0.96 = (x)/(1.414*10^(-3))`

`x = 1.357*10^(-3)m`