A wall has inner and outer surface temperatures of Tsi=14°C and Tso=9°C respectively The interior and exterior air temperatures are T∞i=38°C and T∞o=2°C respectively The inner and outer convection h

We can use the one-dimensional heat conduction equation to solve for the rate of thermal energy storage in the wall:

q = -kA(dT/dx)

where q is the rate of heat transfer, k is the thermal conductivity of the wall, A is the cross-sectional area of the wall, and (dT/dx) is the temperature gradient across the wall.

Assuming steady-state conditions, the temperature gradient across the wall can be calculated as:

(dT/dx) = (T∞,i - Ts,i) / (x/L)

where x is the distance from the inner surface of the wall and L is the thickness of the wall.

Similarly, the rate of heat transfer through convection from the inner surface of the wall can be calculated as:

qconv,i = hiA(T∞,i - Ts,i)

And the rate of heat transfer through convection from the outer surface of the wall can be calculated as:

qconv,o = hoA(Ts,o - T∞,o)

Since the wall is assumed to be in steady-state, the rate of thermal energy storage in the wall must be equal to the sum of the rates of heat transfer through convection:

q = qconv,i + qconv,o

Substituting in the given values, we get:

(dT/dx) = (38°C - 14°C) / (0.2 m) = 120°C/m qconv,i = 33 W/m2-K * 0.2 m2 * (38°C - 14°C) = 132 W qconv,o = 623 W/m2-K * 0.2 m2 * (9°C - 2°C) = 872 W q = 132 W + 872 W = 1004 W

Therefore, the wall is accumulating thermal energy at a rate of 1004 W