The temperature drop through each layer of a | vedclass

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(A) For a steady-state heat flow through a series combination of layers,the rate of heat flow $\frac{dQ}{dt}$ is constant through both layers.The rate

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$k_1 > k_2$

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(A) For a steady-state heat flow through a series combination of layers,the rate of heat flow $\frac{dQ}{dt}$ is constant through both layers.The rate of heat flow is given by $\frac{dQ}{dt} = k A \frac{\Delta T}{\Delta x}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,and $\frac{\Delta T}{\Delta x}$ is the temperature gradient (slope of the $T-x$ graph).Since $\frac{dQ}{dt}$ and $A$ are constant,we have $\frac{\Delta T}{\Delta x} \propto \frac{1}{k}$.From the graph,the slope of the temperature profile in layer $(1)$ is less than the slope in layer $(2)$,i.e.,$|\frac{\Delta T}{\Delta x}|_1 Since the slope is inversely proportional to $k$,a smaller slope implies a larger thermal conductivity.Therefore,$k_1 > k_2$.

The temperature drop through each layer of a two-layer furnace wall is shown in the figure. Assume that the external temperatures $T_1$ and $T_3$ are maintained constant and $T_1 > T_3$. If the thicknesses of the layers $x_1$ and $x_2$ are the same,which of the following statements is correct?

Similar Questions

$A$ rod of length $10 \, cm$ has a cross-sectional area of $100 \, cm^2$ and a thermal conductivity of $400 \, W/m^\circ C$. If the heat flow through the rod is $4000 \, J/s$,find the temperature difference between the two ends in $^\circ C$.

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ $(T_1 > T_2)$. The rate of heat transfer,$\frac{dQ}{dt}$,through the rod in a steady state is given by:

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