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Firdavs [7]
2 years ago
13

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.

Chemistry
1 answer:
Vlad [161]2 years ago
5 0

Answer:

The depth of the damage is given as x   = 1.357*10^{-3}m

Explanation:

From Fick's second law

                 \frac{C_{x}-C_{0}}{C_{s}-C_{0}} = 1 -erf [\frac{x}{2\sqrt{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

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

                  erf[\frac{x}{1.414*10^{-3}} ] = 0.826\\

       Let assume that  z =\frac{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

                  \frac{z-0.95}{0.8260-0.8209}  = 1        

                     z = 0.96

 Substituting the value of z into equation 2

           0.96 = \frac{x}{1.414*10^{-3}}

             x   = 1.357*10^{-3}m

               

 

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