Question 148: Thermodynamics & KTG

Question

Two rectangular blocks, having identical dimensions, can be arranged in either configuration-I or configuration-II as shown in the figure. One of the blocks has thermal conductivity $$\kappa $$ and the other 2$$\kappa $$. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in configuration-I. The time to transport the same amount of heat in configuration-II is

Accepted Answer

Let L and A be length and area of cross-section of each block.
$$	herefore$$ Thermal resistance of block 1 is
$$	herefore$$ $${R_1} = {L \over {\kappa A}}$$
Thermal resistance of block 2 is
$${R_2} = {L \over {2\kappa A}}$$
In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is
$${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$$ ...... (i)
$$	herefore$$ Rate of heat flow in configuration I is
$$	herefore$$ $${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$$ ..... (ii)
In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is
$${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$$ ...... (iii)
$${R_P} = {1 \over 3}{L \over {\kappa A}}$$
$$	herefore$$ Rate of same amount of heat flow in configuration II is
$$	herefore$$ $${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$$ ...... (iv)
Divide (ii) by (iv), we get
$${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} 	imes {{2\kappa A} \over {3L}} = {2 \over 9}$$ (Using (i) and (iii))
$$t' = {2 \over 9}t = {2 \over 9} 	imes 9s = 2s$$

Answer

2.0 s

Answer

3.0 s

Answer

4.5 s

Answer

6.0 s

Thermodynamics & KTG

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