Two solid slabs, P and Q , having the same cross-sectional area are placed in series to form a composite wall. The thickness of slab P is twice the thickness of slab Q , and the thermal conductivity of slab P is half the thermal conductivity of slab Q . In steady state, if the total temperature difference across the entire composite wall is 100^ C , what is the temperature drop across slab P ?

Step-by-step solution

Let the thickness of slab Q be x and its thermal conductivity be k . The thickness of slab P is 2x and its thermal conductivity is k 2 . The thermal resistance of a slab is given by R = L kA . For slab Q , R_Q = x kA . For slab P , R_P = 2x ( k 2 )A = 4x kA = 4R_Q . In steady state, the heat current H is the same through both slabs. The temperature drop across a slab is T = H R . Therefore, T_P = 4 T_Q . Given the total temperature difference is 100^ C : T_P + T_Q = 100^ C T_P + T_P 4 = 100^ C 5 T_P 4 = 100^ C T_P