Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 33

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Question 1
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Four identical rods of same material are joined at ends so as to form a square. If the temperature difference at the ends of a diagonal is 100$$^{o}$$C, then the temperature difference across the ends of another diagonal will be

A
$$0^{o}C$$

B
$$25^{o}C$$

C
$$50^{o}C$$

D
$$100^{o}C$$

Solution

Four identical rods A, B, C, D which having AC & BD as a diagonal point.

Let the temperature at $${A}={T}^{0}C$$

Therefore, temperature at $${C}={t+100}^{0}C$$

Heat is conducted from A to C via B or D. B & D lies midway along with the passing of heat conduction through ABC and ADC respectively.

Hence, temperature of $${B}=\dfrac{t+(t+100)}{2}={t+50}^{0}C$$

similarly,temperature of $${D}=\dfrac{t+(t+100)}{2}={t+50}^{0}C$$

So, the difference of temperature between diagonal $${BD}={0}$$
Question 2
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A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m$$^{2}$$ and thickness $$0.01$$m separated by $$0.05$$m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 27$$^{o}$$C and 0$$^{o}$$C respectively.The thermal conductivity of glass is $$0.8$$ and of air 0.08Wm$$^{-1}$$K$$^{-1}$$

The temperature of the outer glass-air interface is

A
$$26.5^{o}C$$

B
$$25.5^{o}C$$

C
$$24.5^{o}C$$

D
$$23.5^{o}C$$

Solution

$${T}_{2}$$ = temperature of glass-1 and air interface
$${T}_{3}$$ = temperature of air and glass-2 interface

The equation of heat flow are:

$$\displaystyle \frac { { Q }_{ 1 } }{ t } =\frac { { K }_{ 1 }A({ T }_{ 1 }-{ T }_{ 2 }) }{ { d }_{ 1 } }=\frac { { 0.8 }{\times}{1}({ 27}-{ T }_{ 2 }) }{ 0.01 }$$

$$\displaystyle \frac { { Q }_{ 2 } }{ t } =\frac { { K }_{ 2 }A({ T }_{ 2 }-{ T }_{ 3 }) }{ { d }_{ 2} }=\frac { { 0.08 }{\times}{1}({T}_{2}-{ T }_{ 3 }) }{ 0.05 }$$

$$\displaystyle \frac { { Q }_{ 3 } }{ t } =\frac { { K }_{ 3 }A({ T }_{ 3 }-{ T }_{ 4 }) }{ { d }_{ 3 } }=\frac { { 0.8 }{\times}{1}({T}_{3}) }{ 0.01 }$$

In steady state: $$\displaystyle \frac { { q }_{ 1 } }{ t }=\frac { { q }_{ 2 } }{ t }=\frac { { q }_{ 3 } }{ t }$$

On equating $$({1}) ({2})$$ and $${1} {3}$$,we find the following:
$${1350}={51{T}_{2}}-{T}_{3}$$----------(a)
$${27}-{T}_{2}={T}_{3}$$----------(b)

The temperature of outer glass interface $${T}_{2}={26.48}^{0}C$$
Question 3
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Three cylindrical rods $$A, B, C$$ of equal lengths and equal diameters are joined in series as shown in the figure. Their thermal conductivities are $$2k, k, 0.5k$$ respectively. In the steady state, the free ends of rods $$A$$ and $$C$$ are at 100$$^{o}$$C and 0$$^{o}$$C respectively. Neglect loss of heat from the curved surfaces of rods.
The temperature of the junction between rods $$A$$ and $$B$$ is :

A
$$55.7^{o}C$$

B
$$65.7^{o}C$$

C
$$75.7^{o}C$$

D
$$85.7^{o}C$$

Solution

Let temperature between $${A}$$ and $${B}={T}_{1}$$ and temperature between $${B}$$ and $${C}={T}_{2}$$
$$\dfrac { Q }{ t } =\dfrac { KA({ T }_{ 1 }-{ T }_{ 2 }) }{ L } $$
So $$\dfrac { { K }_{ A }A({ 100}-{ T }_{ 1 }) }{ l } =\dfrac { { K }_{ B }A({ T }_{ 1 }-{ T }_{ 2 }) }{ l } =\dfrac { { K }_{ C }A({ T }_{ 2 }-{ 0 }) }{ l } $$
Given that: $${K}_{A}={2K}$$,$${K}_{B}={K}$$,$${K}_{C}={0.5K}$$
$${200}-{2}{{T}_{1}}={T}_{1}-{T}_{2}={0.5}{{T}_{2}}$$

We find: $${T}_{1}=85.71 ^0C$$
$${T}_{2}=57.14 ^0C$$
Question 4
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A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m$$^{2}$$ and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constanttemperatures of 27$$^{o}$$C and 0$$^{o}$$C respectively.The thermal conductivity of glass is 0.8 and of air 0.08Wm$$^{-1}$$K$$^{-1}$$

The temperatures of the inner glass-air interface is

A
$$2.5^{o}C$$

B
$$2.0^{o}C$$

C
$$1.5^{o}C$$

D
$$0.5^{o}C$$

Solution

$${T}_{2}$$ = temperature of glass-1 and air interface
$${T}_{3}$$ = temperature of air and glass-2 interface

The equation of heat flow are:

$$\displaystyle \frac { { Q }_{ 1 } }{ t } =\frac { { K }_{ 1 }A({ T }_{ 1 }-{ T }_{ 2 }) }{ { d }_{ 1 } }=\frac { { 0.8 }{\times}{1}({ 27}-{ T }_{ 2 }) }{ 0.01 }$$

$$\displaystyle \frac { { Q }_{ 2 } }{ t } =\frac { { K }_{ 2 }A({ T }_{ 2 }-{ T }_{ 3 }) }{ { d }_{ 2} }=\frac { { 0.08 }{\times}{1}({T}_{2}-{ T }_{ 3 }) }{ 0.05 }$$

$$\displaystyle \frac { { Q }_{ 3 } }{ t } =\frac { { K }_{ 3 }A({ T }_{ 3 }-{ T }_{ 4 }) }{ { d }_{ 3 } }=\frac { { 0.8 }{\times}{1}({T}_{3}) }{ 0.01 }$$

In steady state: $$\displaystyle \frac { { q }_{ 1 } }{ t }=\frac { { q }_{ 2 } }{ t }=\frac { { q }_{ 3 } }{ t }$$

On equating $$({1}) ({2})$$ and $${1} {3}$$, we find the following :
$${1350}={51{T}_{2}}-{T}_{3}$$----------(a)
$${27}-{T}_{2}={T}_{3}$$----------(b)

The temperature of outer glass interface, $${T}_{2}={26.48}^{0}C$$
So, from equation (b), we find temperature of inner glass interface $${T}_{3}={27}-{{T}_{2}}={27}-{26.48}={0.52}^{0}C$$
Question 5
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Three rods A,B and C have the same dimensions.Their conductivities are $$K_{A}, K_{B}$$ and $$ K_{C}$$ respectively. A and B are placed end to end, with their free ends kept at certain temperature difference. C is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate $$K_{c}$$ must be equal to

A
$$K_{A}+K_{B}$$

B
$$\dfrac{K_{A}+K_{B}}{K_{A}K_{B}}$$

C
$$\dfrac{1}{2}(K_{A}+K_{B})$$

D
$$\dfrac{2K_{A}K_{B}}{K_{A}+K_{B}}$$

Solution

We know that $$\dfrac { Q }{ t } =\dfrac { KA(\Delta T) }{ L } $$
so all rods having same dimensions,therefor for rod $${A} and {B}$$
equivalent thermal conductivity is given as $${ K }_{ equivalent }=\dfrac { { K }_{ A }+{ K }_{ B } }{ { K }_{ A }{ K }_{ B } } $$
$$\dfrac { Q }{ t } =\dfrac { 2{ k }_{ a }{ k }_{ b } }{ ({ k }_{ a }+{ k }_{ b }) } \times \dfrac { A({ T }_{ A }-{ T }_{ B }) }{ 2L } $$
now for rod $${C}$$ we say $$\dfrac { Q }{ t } =\dfrac { {K}_{c}A({ T }_{ A }-{ T }_{ B }) }{ L } $$,here it is mention that both are at same temperature difference.
so equating both of them, we find $${ K }_{ C }=\dfrac { 2{ K}_{ a }{ K }_{ b } }{ ({ K }_{ a }+{ K }_{ b }) } $$
Question 6
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A vessel contains air and saturated vapor. The pressure of air is $$\mathrm{p}_{2}$$ and $$\mathrm{p}_{1}$$ is the S.V. P. On compressing the mixture to one-fourth of its original volume, what is the increase in pressure of the mixture?

A
$$2\mathrm{p}_{1}$$

B
$$2\mathrm{p}_{2}$$

C
$$3\mathrm{p}_{1}$$

D
$$3\mathrm{p}_{2}$$

Solution

Initially total pressure $$=(P_{1}+P_{2})$$
When compressed to $$\dfrac{1}{4}$$ of its volume, the pressure of air becomes $$4P_{2}.$$
Pressure of saturated vapour remains the same.
Total pressure after compression $$=4P_{2}+P_{1}$$
Difference in pressure$$=(4P_{2}+P_{1})-(P_{1}+P_{2})$$
$$=3P_{2}$$
Question 7
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An electric kettle takes 4 A current at 220 V.
How much time will it take to boil 1 kg of water
from temperature
$$ 20^{0}C ?$$ The temperature of
boiling water is
$$ 100^{0}C ?$$

A
12.6 min

B
4.2min

C
6.3 min

D
8.4 min

Solution

Heat required to boil water
$$Q = mc \Delta T$$
$$= 1 \times 1 \times (100 20) = 80kcal$$
Heat given by supply
$$H = V.i.t = 220 \times 4 \times t$$
$$H = Q$$
$$ \Rightarrow 220 \times 4 \times t = 80,000$$
$$t=\frac{80,000}{220 \times 4}$$
$$=\frac{1000}{11}sec$$
$$=\frac{1000}{11 \times 60}=1.5 min$$
Question 8
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Consider two rods of same length and different specific heats (S$$_{1}$$, S$$_{2}$$), conductivity (K$$_{1}$$, K$$_{2}$$) and area of cross-sections (A$$_{1}$$, A$$_{2}$$) and both having temperature T$$_{1}$$ and T$$_{2}$$ at their ends. If rate of loss of heat due to conduction is 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
$$\frac{K_{2}A_{1}}{S_{2}}=\frac{K_{1}A_{2}}{S_{1}}$$

Solution

Given:
For rod 1
Cross-sectional area $$=A_1$$
Specific Heat $$=S_1$$
Temperaturedifference between the ends of the rod $$=T_1-T_2$$
Thermal Conductivity $$= K_1$$

For rod 2:
Cross-sectional area $$=A_2$$
Specific Heat $$=S_2$$
Temperature difference at ends of the rod $$=T_1-T_2$$
Thermal Conductivity $$= K_2$$

By Fourier's law of heat conduction:
For 1st rod
$$\dfrac{dQ_1}{dt}=-K_1\dfrac{A_1dT}{L}$$

For 2nd rod:
$$\dfrac{dQ_2}{dt}=-K_2\dfrac{A_2dT}{L}$$

The rate of heat transfer is same;
Thus,
$$-K_1\dfrac{A_1dT}{L}=-K_2\dfrac{A_2dT}{L}$$
or
$$K_1A_1=K_2A_2$$
Question 9
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Two plates of same area are placed in contact. Their thickness as well as thermal conductivities are in the ratio 2 : 3. The outer surface of one plate is maintained at $$10^\circ C$$ and that of other at $$0^\circ C$$. What is the temp.of the common surface?

A
$$0^\circ C$$

B
$$2.5^\circ C$$

C
$$5^\circ C$$

D
$$6.5^\circ C$$

Solution

Let $$A$$ be the area of the both the plates.

Given ,

$$ \dfrac{k_1}{k_2}$$ = $$ \dfrac {2}{3} $$ -------$$(i)$$

where $$k_1$$ and $$k_2$$ are the thermal conductivities of plate 1 and 2 respectively.

And

$$ \dfrac{l_1}{l_2}$$ = $$ \dfrac {2}{3} $$ ------- $$(ii)$$

where $$l_1$$ and $$l_2$$ are the thermal conductivities of plate 1 and 2 respectively.

Let the temperature of the common surface be $$T$$

Equating the rate of heat transfers through both the plates

$$ \dfrac{k_1 A (10-T)}{L_1}$$ = $$ \dfrac{k_2 A (T-0)}{L_2}$$

Using $$ (i) $$ and $$ (ii) $$

We have ,

$$ 10-T = T-0 $$

Hence , $$ T = 5^0C $$
Question 10
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A cylinder of radius r and thermal conductivity $$K_1$$ is surrounded by a cylindrical shell of inner radius r and outer radius 2 r, whose thermal conductivity is $$K_2$$. There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temps. $$T_1$$ and $$T_2$$. The effective thermal conductivity of the system, in the steady state is

A
$$\cfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$$

B
$$K_1+K_2$$

C
$$\cfrac{{{K_1} + 3{K_2}}}{4}$$

D
$$\cfrac{{3{K_1} + 3{K_2}}}{4}$$

Solution

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 33

Result

Thermal Properties of Matter Test - 33

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SHARING IS CARING

Question 1

$$0^{o}C$$

$$25^{o}C$$

$$50^{o}C$$

$$100^{o}C$$

Solution

Question 2

$$26.5^{o}C$$

$$25.5^{o}C$$

$$24.5^{o}C$$

$$23.5^{o}C$$

Solution

Question 3

$$55.7^{o}C$$

$$65.7^{o}C$$

$$75.7^{o}C$$

$$85.7^{o}C$$

Solution

Question 4

$$2.5^{o}C$$

$$2.0^{o}C$$

$$1.5^{o}C$$

$$0.5^{o}C$$

Solution

Question 5

$$K_{A}+K_{B}$$

$$\dfrac{K_{A}+K_{B}}{K_{A}K_{B}}$$

$$\dfrac{1}{2}(K_{A}+K_{B})$$

$$\dfrac{2K_{A}K_{B}}{K_{A}+K_{B}}$$

Solution

Question 6

$$2\mathrm{p}_{1}$$

$$2\mathrm{p}_{2}$$

$$3\mathrm{p}_{1}$$

$$3\mathrm{p}_{2}$$

Solution

Question 7

12.6 min

4.2min

6.3 min

8.4 min

Solution

Question 8

$$K_{1}A_{1}=K_{2}A_{2}$$

$$\frac{K_{1}A_{1}}{S_{1}}=\frac{K_{2}A_{2}}{S_{2}}$$

$$K_{2}A_{1}=K_{1}A_{2}$$

$$\frac{K_{2}A_{1}}{S_{2}}=\frac{K_{1}A_{2}}{S_{1}}$$

Solution

Question 9

$$0^\circ C$$

$$2.5^\circ C$$

$$5^\circ C$$

$$6.5^\circ C$$

Solution

Question 10

$$\cfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$$

$$K_1+K_2$$

$$\cfrac{{{K_1} + 3{K_2}}}{4}$$

$$\cfrac{{3{K_1} + 3{K_2}}}{4}$$

Solution

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