Question 39: Thermodynamics & KTG

Question

Rods $A$ and $B$ have their lengths in the ratio $1: 2$. Their thermal conductivities are $K_1$ and $K_2$ respectively. The temperatures at the ends of each rod are $\mathrm{T}_1$ and $\mathrm{T}_2$. If the rate of flow of heat through the rods is equal, the ratio of area of cross section of $A$ to that of $B$ is

Accepted Answer

We use the oneÃ¢ÂÂdimensional Fourier law for steady heat flow
$$\frac{Q}{t} = \frac{K\,A\,(T_1 - T_2)}{L}\,. $$
For rods A and B we have
Ã¢ÂÂ¢ Lengths: $L_A:L_B = 1:2$, so $L_B = 2L_A$.
Ã¢ÂÂ¢ Conductivities: $K_A = K_1$, $K_B = K_2$.
Ã¢ÂÂ¢ Same endÃ¢ÂÂtemperatures $T_1,T_2$, hence the same $\Delta T=T_1 - T_2$.
Ã¢ÂÂ¢ Equal heatÃ¢ÂÂflow rates: $\bigl(Q/t\bigr)_A = \bigl(Q/t\bigr)_B$.
Setting them equal,
$$\frac{K_1\,A_A\,\Delta T}{L_A}   =\frac{K_2\,A_B\,\Delta T}{L_B}   \;\;\Longrightarrow\;\;   \frac{A_A}{A_B}   =\frac{K_2}{K_1}\,\frac{L_A}{L_B}   =\frac{K_2}{K_1}\,\frac{1}{2}   =\frac{K_2}{2\,K_1}\,. $$
Answer: $\displaystyle \frac{A_A}{A_B}=\frac{K_2}{2\,K_1}$.

Answer

$\frac{2 K_2}{K_1}$

Answer

$\frac{K_2}{4 K_1}$

Answer

$\frac{K_2}{K_1}$

Answer

$\frac{K_2}{2 K_1}$

Thermodynamics & KTG

Rods A and B have their lengths in the ratio 1: 2. Their thermal conductivities are K_1 and K_2 respectively. The temperatures at the ends of each rod are \mathrm{T}_1 and \mathrm{T}_2. If the rate of flow of heat through the rods is equal, the ratio of area of cross section of A to that of B is

Choose the correct answer: