Q1: The left face of a one dimensional slab of thickness 0.2 m is maintained at 80°C and the right face is exposed to air at 30°C. The thermal conductivity of the slab is 1.2 W/m.K and the heat transfer co-efficient from the right face is 10 W/m².K. At steady state, the temperature of the right face in °C is

ANS:D - 48.7

To solve this problem, we can apply the steady-state heat conduction equation for a one-dimensional slab: Q=Lk⋅A⋅(T1​−T2​)​ Where:

• Q is the rate of heat transfer through the slab (W),
• k is the thermal conductivity of the slab (W/m·K),
• A is the cross-sectional area of the slab (m²),
• T1​ is the temperature of the left face of the slab (°C),
• T2​ is the temperature of the right face of the slab (°C),
• L is the thickness of the slab (m).

We'll also apply Newton's law of cooling for the right face, where the heat transfer rate is given by: =ℎ⋅⋅(ambient−2)Q=h⋅A⋅(Tambient​−T2​) Where:

• ℎh is the heat transfer coefficient (W/m²·K),
• ambientTambient​ is the ambient temperature (°C).

We can set these two expressions for Q equal to each other since the heat transfer rate through the slab must be equal to the heat transfer rate from the right face: ⋅(1−2)=ℎ⋅⋅(ambient−2)Lk⋅A⋅(T1​−T2​)​=h⋅A⋅(Tambient​−T2​) We can then solve for T2​ using the given values. Given:

• 1=80°T1​=80°C
• ambient=30°Tambient​=30°C
• =1.2 W/m\cdotpKk=1.2W/m\cdotpK
• ℎ=10 W/m²\cdotpKh=10W/m²\cdotpK
• =0.2 mL=0.2m

Let's solve for T2​: First, we calculate the cross-sectional area A using the thickness L of the slab: =1×1=1 m2A=1×1=1m2 Now, we'll set up the equation: 1.2⋅1⋅(80−2)0.2=10⋅1⋅(30−2)0.21.2⋅1⋅(80−T2​)​=10⋅1⋅(30−T2​) 1.2⋅(80−2)0.2=10⋅(30−2)0.21.2⋅(80−T2​)​=10⋅(30−T2​) 6⋅(80−2)=100⋅(30−2)6⋅(80−T2​)=100⋅(30−T2​) 480−62=3000−100 2480−6T2​=3000−100T2​ T2​=2520 T2​=942520​ °T2​≈26.81°C So, the temperature of the right face of the slab at steady state is approximately 26.81°C.