Find the temperature of the bar (made of aluminum, k = 202.4 w/m*k) with a rectangular cross-section at the center point. the left end of rectangular is 80 c, the right end is 40 c and side walls convection boundary layer (steam temperature 22 c) and convection coefficient is 15 w/m^(2)k. length is 10 mm, height 10 mm and width 100 mm. show working out.

Explanation:

As it is given that no heat is generating and the system is present in steady state.

As we know that,

    =

So, according to the given conditions, = 0, and = 0

So,     = 0 , =

               T =

Hence, boundary conditions will be as follows.

At x = 0, T =

At x = L, T =

Applying boundary condition is

Therefore, =

               40 =

                  = -400

Hence, temperature distribution in a block will be as follows.

                  T = -400x + 80

At center, x =

Therefore,  T =

                      =

Thus, we can conclude that the temperature of block at the center point is .

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3