Rate of conduction of heat is given as tΔQ=LKA(T2−T1) According to the question ΔtΔQ1=ΔtΔQ2⇒LK1A1(T1−T2)=LK1A1(T1−T2)⇒K1A1=K2A2
Consider two rods of the same length and different specific heats (S1,S2), conductivities (K1,K2) and area of cross sections (A1,A2) and both having temperature T1 and T2 at their ends. If the rate of loss of heat due to conduction is equal then:
Held on 30 Apr 2002 · Verified 9 Jul 2026.
K1A1=K2A2
S2K1A1=S1K2A2
K2A1=K1A2
S2K2A1=S1K1A2
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