1.
The figure shows a system of two concentric spheres of radii $${{r_1}}$$ and $${{r_2}}$$ are kept at temperatures $${{T_1}}$$ and $${{T_2}},$$ respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
A.
$$ln\left( {\frac{{{r_2}}}{{{r_1}}}} \right)$$
B.
$$\frac{{\left( {{r_2} - {r_1}} \right)}}{{\left( {{r_1}{r_2}} \right)}}$$
C.
$${\left( {{r_2} - {r_1}} \right)}$$
D.
$$\frac{{{r_1}{r_2}}}{{\left( {{r_2} - {r_1}} \right)}}$$
Consider a shell of thickness $$(dr)$$ and of radius $$(r)$$ and the temperature of inner and outer surfaces of this shell be $$T,\left( {T - dT} \right)$$
$$\eqalign{
& H = \frac{{KA\left[ {\left( {T - dT} \right) - T} \right]}}{{dr}} = \frac{{ - KAdT}}{{dr}} \cr
& H = - 4\,\pi K{r^2}\frac{{dT}}{{dr}}\,\,\left( {\because \,\,A = 4\,\pi {r^2}} \right) \cr
& {\text{Then, }}\left( H \right)\int\limits_{{r_1}}^{{r_2}} {\frac{1}{{{r^2}}}dr = - 4\,\pi K\int\limits_{{T_1}}^{{T_2}} {dT} } \cr
& H\left[ {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right] = - 4\,\pi K\left[ {{T_2} - {T_1}} \right] \cr
& {\text{or, }}H = \frac{{ - 4\,\pi K{r_1}{r_2}\left( {{T_2} - {T_1}} \right)}}{{\left( {{r_2} - {r_1}} \right)}} \cr} $$
2.
Which of the following circular rods, (given radius $$r$$ and length $$l$$) each made of the same material and whose ends are maintained at the same temperature will conduct most heat ?
As from law of heat transfer through conduction
$$\eqalign{
& H = \frac{{\Delta Q}}{{\Delta t}} = KA\left( {\frac{{{T_1} - {T_2}}}{1}} \right) \cr
& \Rightarrow H \propto \frac{{{r^2}}}{l}\,......\left( {\text{i}} \right) \cr} $$
(A) When $$r = 2{r_0};l = 2{l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{2{l_0}}} \Rightarrow H \propto \frac{{2r_0^2}}{{{l_0}}}$$
(B) When $$r = 2{r_0};l = {l_0}$$
$$H \propto \frac{{{{\left( {2{r_0}} \right)}^2}}}{{{l_0}}} \Rightarrow H \propto \frac{{4r_0^2}}{{{l_0}}}$$
(C) When $$r = {r_0};l = {l_0} \Rightarrow H \propto \frac{{r_0^2}}{{{l_0}}}$$
(D) When $$r = {r_0};l = 2{l_0} \Rightarrow H \propto \frac{{r_0^2}}{{2{l_0}}}$$
It is obvious that heat conduction will be more in case (B).
3.
The two ends of a rod of length $$L$$ and a uniform cross-sectional area $$A$$ are kept at two temperatures $${T_1}$$ and $${T_2}\left( {{T_1} > {T_2}} \right).$$ The rate of heat transfer, $$\frac{{dQ}}{{dt}},$$ through the rod in a steady state is given by
A.
$$\frac{{dQ}}{{dt}} = \frac{{KL\left( {{T_1} - {T_2}} \right)}}{A}$$
B.
$$\frac{{dQ}}{{dt}} = \frac{{K\left( {{T_1} - {T_2}} \right)}}{{LA}}$$
C.
$$\frac{{dQ}}{{dt}} = KLA\left( {{T_1} - {T_2}} \right)$$
D.
$$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$
For a rod of length $$L$$ and area of cross-section $$A$$ whose faces are maintained at temperatures $${T_1}$$ and $${T_2}$$ respectively.
Then in steady state the rate of heat flowing from one face to the other face in time $$t$$ is given by $$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$
4.
A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $$\theta $$ along the length $$x$$ of the bar from its hot end is best described by which of the following figures?
The heat flow rate is given by
$$\eqalign{
& \frac{{dQ}}{{dt}} = \frac{{kA\left( {{\theta _1} - \theta } \right)}}{x} \cr
& \Rightarrow \,\,{\theta _1} - \theta = \frac{x}{{kA}}\frac{{dQ}}{{dt}} \cr
& \Rightarrow \,\,\theta = {\theta _1} - \frac{x}{{kA}}\frac{{dQ}}{{dt}} \cr} $$
where $${\theta _1}$$ is the temperature of hot end and $${\theta}$$ is
temperature at a distance $$x$$ from hot end.
The above equation can be graphically represented by
option (A) .
5.
Two rods $$A$$ and $$B$$ of different materials are welded together as shown in figure. Their thermal conductivities are $${K_1}$$ and $${K_2}.$$ The thermal conductivity of the composite rod will be :
A.
$$\frac{{3\left( {{K_1} + {K_2}} \right)}}{2}$$
Heat current $$H = {H_1} + {H_2}$$
$$\eqalign{
& = \frac{{{K_1}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} + \frac{{{K_2}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} \cr
& \frac{{{K_{EQ}}2A\left( {\;{T_1} - {T_2}} \right)}}{d} = \frac{{A\left( {{T_1} - {T_2}} \right)}}{d}\left[ {{K_1} + {K_2}} \right] \cr} $$
Hence equivalent thermal conductivities for two rods of equal area is given by
$${K_{EQ}} = \frac{{{k_1} + {k_2}}}{2}$$
6.
The temperature of the two outer surfaces of a composite slab, consisting of two materials having co-efficients of thermal conductivity $$K$$ and $$2\,K$$ and thickness $$x$$ and $$4\,x,$$ respectively, are $${T_2}$$ and $${T_1}\left( {{T_2} > {T_1}} \right).$$ The rate of heat transfer through the slab, in a steady state is $$\left( {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right)f.$$ with $$f$$ equal to
7.
Consider two rods of same length and different specific heats $$\left( {{s_1},{s_2}} \right),$$ thermal conductivities $$\left( {{K_1},{K_2}} \right)$$ and areas of cross-section $$\left( {{A_1},{A_2}} \right)$$ and both having temperatures $$\left( {{T_1},{T_2}} \right)$$ at their ends. If their rate of loss of heat due to conduction are equal, then
A.
$${K_1}{A_1} = {K_2}{A_2}$$
B.
$$\frac{{{K_1}{A_1}}}{{{s_1}}} = \frac{{{K_2}{A_2}}}{{{s_2}}}$$
C.
$${K_2}{A_1} = {K_1}{A_2}$$
D.
$$\frac{{{K_2}{A_1}}}{{{s_2}}} = \frac{{{K_1}{A_2}}}{{{s_1}}}$$
Rate of loss of heat by conduction is, $$H = \frac{{\Delta Q}}{{\Delta t}} = KA\left( {\frac{{{T_1} - {T_2}}}{l}} \right)$$
All the symbols have their usual meaning.
For first rod, $${H_1} = {K_1}{A_1}\left( {\frac{{{T_1} - {T_2}}}{{{l_1}}}} \right)$$
For second rod, $${H_2} = {K_2}{A_2}\left( {\frac{{{T_1} - {T_2}}}{{{l_2}}}} \right)$$
but $${l_1} = {l_2}$$ i.e. of same length
and $${H_1} = {H_2}$$ i.e. same rate of loss of heat through conduction
So, we have
$${K_1}{A_1}\left( {{T_1} - {T_2}} \right) = {K_2}{A_2}\left( {{T_1} - {T_2}} \right)$$
or $${K_1}{A_1} = {K_2}{A_2}$$ NOTE
Heat transfer occurs only between regions that are at different temperatures, and the rate of heat flow is $$\frac{{dQ}}{{dt}}.$$
8.
Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles triangle $$ABC,$$ right-angled at $$B.$$ The points $$A$$ and $$B$$ are maintained at temperatures $$T$$ and $$\left( {\sqrt 2 } \right)$$ $$T$$ respectively. In the steady state, the temperature of the point $$C$$ is $${T_c}.$$ Assuming that only heat conduction takes place, $$\frac{{{T_c}}}{T}$$ is
A.
$$\frac{1}{{2\left( {\sqrt 2 - 1} \right)}}$$
B.
$$\frac{3}{{\sqrt 2 + 1}}$$
C.
$$\frac{1}{{\sqrt 3 \left( {\sqrt 2 - 1} \right)}}$$
Heat flow from $$B$$ to $$A,A$$ to $$C$$ and $$C$$ to $$B$$ (for steady state condition, $$\frac{{\Delta Q}}{{\Delta t}}$$ is same)
Where $$\frac{{\Delta Q}}{{\Delta t}} = \frac{{kA\Delta T}}{\ell }$$
For sides $$AC$$ and $$CB{\left( {\frac{{\Delta T}}{{\sqrt 2 a}}} \right)_{AC}} = {\left( {\frac{{\Delta T}}{a}} \right)_{CB}}$$
$$\eqalign{
& \Rightarrow \frac{{T - {T_c}}}{{\sqrt 2 a}} = \frac{{{T_c} - \sqrt 2 T}}{a} \cr
& \Rightarrow T - {T_c} = \sqrt 2 {T_c} - 2T \cr
& \Rightarrow 3T = {T_c}\left( {\sqrt 2 + 1} \right) \Rightarrow \frac{{{T_c}}}{T} = \frac{3}{{\sqrt 2 + 1}} \cr} $$
9.
A slab of stone of area $$0.36\,{m^2}$$ and thickness $$0.1\,m$$ is exposed on the lower surface to steam at $${100^ \circ }C.$$ A block of ice at $${0^ \circ }C$$ rests on the upper surface of the slab. In one hour $$4.8\,kg$$ of ice is melted. The thermal conductivity of slab is : (Given latent heat of fusion of ice $$ = 3.36 \times {10^5}\,J\,k{g^{ - 1}}.$$ ):
(A) Conduction is the process of transmission of heat in a body from the hotter part to the colder part without any bodily movement of constituent atoms or molecules of the body.
(B) In convection, the heated lighter particles move upward and colder heavier particles move downward to their place. This depends on weight and hence, on gravity.
(C) Radiation is the process of transmission of heat from one body to another body through electromagnetic waves even through vacuum, irrespective of their temperatures. Hence, choice (B) is correct.