In the first experiment, two identical conducting rods are joined one after the other and this combination is connected to two vessels, one containing water at 100°C and the other containing ice at 0°C (see  Fig.). In the second experiment; the two rods are placed one on top of the other and connected to the same vessels. If q₁ and q₂ (in gram per second) are the respective rates of melting of ice in the two cases, then the ratio $\frac{ q_1}{q_2}$ is 

 

If a steady temperature difference (θ₁-θ₂)  is maintained between the ends of a conducting rod of length Land cross-sectional area A, the rate of flow of heat through the rod is given by 

$q=\frac{ka({\theta }_1-{\theta }_2)}{L}$

where k is the coefficient of thermal conductivity of the material of the rod. In the first experiment, the two rods are connected in series. If two rods of equal cross-sectional areas and of lengths L₁ and L₂ and conductivities k₁ and k₂ are joined in series, the equivalent conductivity k_s is given by

$\frac{L_1+L_2}{k_s}=\frac{L_1}{k_1}+ \frac{L_2}{k_2} \left(1\right)$

For two identical rods, L₁ = L₂ = La nd k₁ = k₂ = k, in which case, Eq. (1) gives k_s = k. Further, when two identical rods are joined in series, the length of the composite rod is (2L) but its cross-sectional area is A, the same as that of each rod. Hence the rate of flow
of heat in this case is given by

$q_1=\frac{kA({\theta }_1-{\theta }_2)}{\left(2L\right)} \left(2\right)$

In the second case, the two rods are connected in parallel. If two rods of equal lengths and equal crosssection areas and having conductivities k₁ and k₂ are joined in parallel, the equivalent conductivity of the composite rod is given by

k_P = k₁ + k₂

For two identical rods,k₁ = k₂ = k. Hence k_P = (2k). Furthermore, the cross-sectional area of the composite rod is (2A). Therefore, in the second case, the rate of flow of heat is given by

$q_2=\frac{\left(2k\right)\left(2A\right)({\theta }_1-{\theta }_2)}{\left(L\right)} \left(3\right)$

Dividing (2),by (3), we get $\frac{ q_1}{q_2}$=$\frac{1}{8}$ . Now, the rate of melting of ice is proportional to the rate of flow of heat.