ANS:C - 70

To solve this problem, we can use the principle of thermal resistance across the interface between materials B and C. The rate of heat transfer (Q) through a composite wall consisting of multiple layers is given by Fourier's Law: Q=∑RΔT​ Where:

• ΔT is the temperature difference across the wall.
• ∑R is the sum of the thermal resistances of all layers.

The thermal resistance (R) of a material is calculated as the thickness (d) divided by the product of the material's thermal conductivity (k) and its area (A): R=k⋅Ad​ Given:

• Material A: =20kA​=20 kcal/hr.m.°C
• Material B: =40kB​=40 kcal/hr.m.°C
• Material C: =60kC​=60 kcal/hr.m.°C
• Thickness of each material (let's assume equal thickness for simplicity): dA​=dB​=dC​=d
• Temperature outside A: A=30°TA​=30°C
• Temperature outside C: C=100°TC​=100°C

We need to find the temperature at the interface between B and C (interfaceTinterface​). First, let's calculate the thermal resistances of each material: RA​=kA​d​ RB​=kB​d​ RC​=kC​d​ Then, the total thermal resistance (∑R) across the composite wall is the sum of the resistances of the individual layers: ∑R=RA​+RB​+RC​ We know that the temperature difference across the wall (ΔT) is the difference between the temperatures outside A and C: AΔT=TC​−TA​ Now, we can use Fourier's Law to find interfaceTinterface​: Tinterface​=TA​+∑RRA​​⋅ΔT Let's calculate: 20RA​=kA​d​=20d​ 40RB​=kB​d​=40d​ 60RC​=kC​d​=60d​ 60=660+360+260=1160∑R=RA​+RB​+RC​=20d​+40d​+60d​=606d​+603d​+602d​=6011d​ ΔC−A=100−30=70°ΔT=TC​−TA​=100−30=70°C interface=30+201160⋅70Tinterface​=30+6011d​20d​​⋅70 interface=30+311⋅70Tinterface​=30+113​⋅70 interface=30+21011Tinterface​=30+11210​ interface≈30+19.09Tinterface​≈30+19.09 interface≈49.09°Tinterface​≈49.09°C So, the interface between B and C will be at approximately 49.09°C. Therefore, the closest option is 50°50°C.