Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 63

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Weekly Quiz Competition

Question 1
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Keeping the number of moles, volume and pressure the same, which of the following are the same for all ideal gas?

A
rms speed of a molecule

B
density

C
temperature

D
average of magnitude of momentum

Solution

Ideal gas equation $$\rightarrow PV=nRT$$
Given that $$n_1V_1P$$ are same.
So, $$T$$ i.e, temperature is same for all ideal gas.
Hence, the answer is Temperature.
Question 2
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A metal ball immersed in water weighs $$w_1$$ at $$5^{\circ}C\;and\;w_2$$ at $$50^{\circ}C$$. The coefficient of cubical expansion of metal is less than that of water. Then

A
$$w_1\,>\,w_2$$

B
$$w_1\,<\,w_2$$

C
$$w_1=w_2$$

D
Insufficient information

Solution

Weight of immersed ball = mg -buoyancy force
Buoyancy force is equal to weight of liquid displaced.
now when we increase the temperature both will expand but expansion in water will be more,hence the mass of volume displaced will decrease.
Hence, $$w_1<w_2$$.
Question 3
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A black body is at $$(727)^{0}C$$. It emits energy at a rate which is proportional to:

A
$$(727)^{4}$$

B
$$(727)^{2}$$

C
$$(1000)^{4}$$

D
$$(1000)^{2}$$

Solution

$$E= \sigma T^{4}\therefore E \infty (727+273)^{4}\Rightarrow E \infty \left ( 1000 \right )^{4}$$
Question 4
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Three rods $$A,B$$ and $$C$$ have the same dimensions. Their thermal conductivities $$k_A, \space k_B$$ and $$k_C$$ respectively. $$A$$ and $$B$$ are placed end to end, with the free ends kept at a certain temperature difference. $$C$$ is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate $$k_C$$ equal to

A
$$k_A+k_B$$

B
$$\displaystyle \frac{k_Ak_B}{k_A+k_B}$$

C
$$\displaystyle \frac{1}{2}(k_A+k_B)$$

D
$$\displaystyle 2\frac{k_Ak_B}{k_A+k_B}$$

Solution

Given that three rods A, B and C have same dimensions. Their thermal conductivities $$k_{A}$$, $$k_{B}$$ and $$k_{C}$$ respectively. A and B are placed end to end,
with the free ends kept at a certain temperature difference. C is placed separately with its ends kept at same temperature difference.
Heat flow through A and B rods at thermal equilibrium is equal
$${ H }_{ A }={ H }_{ B }=\dfrac { { k }_{ A }a(T-{ T }_{ 1 }) }{ L } =\dfrac { { k }_{ B }a({ T }_{ 2 }-T) }{ L } \\ \quad \quad \quad \quad \quad ({ k }_{ A }+{ k }_{
B })T=\quad { k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad T\quad =\dfrac { { k }_{ A }{ T }_{ 1 }+{
k }_{ B }{ T }_{ 2 } }{ { k }_{ A }+{ k }_{ B } } $$
where T is the temperature of the common end Of A and B, $$T_{1}$$ and $$T_{2}$$ are temperatures at the other ends of A and B, $$a$$ is area and $$L$$ is the length of the each rod
Heat flow through C
$${ H }_{ C }=\dfrac { { k }_{ C }a({ T }_{ 2 }-{ T }_{ 1 }) }{ L } $$
If the rate of heat flow for the above two arrangements are equal
$${ H }_{ A }={ H }_{ B }={ H }_{ C }$$
$$\dfrac { { k }_{ A }a(\dfrac { { k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 } }{ { k }_{ A }+{ k }_{ B } } -{ T }_{ 1 }) }{ L } =\dfrac { { k }_{ C }a({ T }_{ 2 }-{ T }
_{ 1 }) }{ L } \\ \dfrac { { k }_{ A }({ k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 }-{ k }_{ A }{ T }_{ 1 }-{ k }_{ B }{ T }_{ 1 }) }{ { k }_{ A }+{ k }_{ B } } ={ k }
_{ C }({ T }_{ 2 }-{ T }_{ 1 })\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \dfrac { { k }_{ A }{ k }_{ B } }{ { k }_{ A }+{ k }_{ B } } ={ k }_{ C }$$
option (B) is the correct answer.
Question 5
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An insulated cylindrical tube of an air-conditioner's condenser contains a hot fluid. Temperature of fluid is $${500}^{o}C$$ and outside temperature is $${40}^{o}C$$. Hot fluid tube is very thin and is covered with $$3$$ layers of different insulating materials. Cross-section of the tube is as shown in figure. Given $${ r }_{ 1 }=1cm;{ r }_{ 2 }=2cm;{ r }_{ 3 }=8cm;{ r }_{ 4 }=64cm$$
$${ K }_{ A }=1W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 };{ K }_{ B }=2W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 };{ K }_{ C }=3W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 }$$
Heat loss per unit length (in watts) of tube will be:

A
$$\cfrac { 460 }{ \ln { (2) } } $$

B
$$\cfrac { 460 }{ \pi \ln { (2) } } $$

C
$$\cfrac { \pi \times 460 }{ \ln { (2) } } $$

D
$$\cfrac {2 \pi \times 460 }{ 3\ln { (2) } } $$

Solution

To solve this question let's have some general discussion.
Top view of cylinder only one layer is A (fig i)
Thermal resistance $$\dfrac{l}{KA}$$
Thermal resistance of element with thickness dr of radius r
$$\int { dR=\int _{ { r }_{ 1 } }^{ { r }_{ 2 } }{ \dfrac { dr }{ K2\pi rL } } } $$
$$R=\dfrac { 1 }{ 2\pi KL } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) $$ (1)
Now in fig(ii) there are three layers and these layers are in series.
Resistance of layer A using formula(1)
$$=\dfrac { 1 }{ 2\pi { K }_{ A }L } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) $$ (2)
similarly resistance of layer B
$$=\dfrac { 1 }{ 2\pi { K }_{ B }L } ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) $$ (3)
Similarly resistance of layer C
$$=\dfrac { 1 }{ 2\pi { K }_{ C }L } ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) $$
Requivalent $$=R_A+R_B+R_C$$
$$=\dfrac { 1 }{ 2\pi { K }_{ A }L } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) +\dfrac { 1 }{ 2\pi { K }_{ B }L } ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) +\dfrac { 1 }{ 2\pi { K }_{ C }L } ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) $$

$$H=\dfrac { \left( { T }_{ 1 }-{ T }_{ 4 } \right) }{ \dfrac { ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }{ 2\pi { K }_{ A }L } +\dfrac { ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) }{ 2\pi { K }_{ B }L } +\dfrac { ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) }{ 2\pi { K }_{ C }L } } $$

$$H=\dfrac { 2\pi L\times \left( { T }_{ 1 }-{ T }_{ 4 } \right) }{ \dfrac { ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }{ { K }_{ A } } +\dfrac { ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) }{ { K }_{ B } } +\dfrac { ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) }{ { K }_{ C } } } $$

$$=\dfrac { 2\pi L\times 460 }{ \dfrac { { ln }2 }{ 1 } +\dfrac { ln4 }{ 2 } +\dfrac { ln8 }{ 3 } } $$

$$H=\dfrac { 2\pi L\times 460 }{ ln2+ln2+ln2 } $$
$$H=\left( \dfrac { 2 }{ 3 } \right) \dfrac{\pi L\times 460}{ln2}$$W
$$\dfrac{H}{L}=\dfrac{2}{3}\dfrac{\pi \times460}{ln2}$$
Question 6
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The value of $$C_p-C_v$$ is 1.09 R for a gas sample in state A and is 1.00 R in state R. Let $$T_A, T_B$$ denote the temperature and $$p_A$$ and $$p_B$$ denote the pressure of the states A and B respectively. Then

A
$$p_A < p_B$$ and $$T_A > T_B$$

B
$$p_A > p_B$$ and $$T_A > T_B$$

C
$$p_A = p_B$$ and $$T_A < T_B$$

D
$$p_A > p_B$$ and $$T_A < T_B$$

Solution
Question 7
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Two slabs $$A$$ and $$B$$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab $$A$$ is twice that of $$B$$. The coefficient of thermal conductivity or slab $$A$$ is twice that of $$B$$. The first surface of slab $$A$$ is maintained at $$100$$, while the second surface of slab $$B$$ is maintained at $$25$$. The temperature at the contact of their surfaces is

A
$$62.5$$

B
$$45$$

C
$$55$$

D
$$85$$

Solution

The temperature at the contact of the surface
$$=\dfrac { { K }_{ 1 }{ d }_{ 2 }{ \theta }_{ 1 }+{ K }_{ 2 }{ d }_{ 1 }{ \theta }_{ 2 } }{ { K }_{ 1 }{ d }_{ 2 }+{ K }_{ 2 }{ d }_{ 1 } } $$

$$=\dfrac { 2{ K }_{ 2 }{ d }_{ 2 }\times 100+2{ d }_{ 2 }\times { K }_{ 2 }\times 25 }{ 2{ K }_{ 2 }{ d }_{ 2 }+{ K }_{ 2 }2{ d }_{ 2 } } $$

$$=\dfrac { 200+50 }{ 4 } =62.5$$
Question 8
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A thermocol box has a total wall area (including the lid) of $$1.0{ m }^{ 2 }$$ and wall thickness of $$3cm$$. It is filled with ice at $${ 0 }^{ o }C$$. If the average temperature outside the box is $${ 30 }^{ o }C$$ throughout the day, the amount of ice that melts in one day is
(Use $${ K }_{ thermocol }=0.03W/mK,{ L }_{ fusion(ice) }=3.00\times { 10 }^{ 5 }J/kg\quad $$

A
$$1kg$$

B
$$2.88kg$$

C
$$25.92kg$$

D
$$8.64kg$$

Solution

Given : $$A=I{ m }^{ 2 }$$
$$L=3\times { 10 }^{ 5 } \ J/kg$$
$$l=3cm=3\times { 10 }^{ -2 }m$$
$$ t=24\times 60\times 60 \ s$$
$$\Delta \theta =30-0={ 30 }^{ o }C$$
Using $$\cfrac { Q }{ t } =\cfrac { KA }{ l } \Delta \theta \quad $$
or $$\cfrac { mL }{ t } =\cfrac { KA }{ l } \Delta \theta \quad $$
Putting the values, $$\cfrac { m\times 3\times { 10 }^{ 5 } }{ 24\times 60\times 60 } =\cfrac { 0.03\times 1 }{ 3\times { 10 }^{ -2 } } \times 30$$
$$\implies \ m=8.641kg$$
Question 9
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A shining metallic ball with a small black spot on its surface is heated to a very high temperature and then quickly taken to a dark room. Then:

A
Both appear equally bright

B
The spot appears brighter than the ball

C
The spot appears darker than the ball

D
Both are invisible in the dark room

Solution
Question 10
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A rod of length $$l$$ with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as $$k = \alpha/ T$$, where $$\alpha$$ is a constant. The ends of the rod are kept at temperature $$T_{1}$$ and $$T_{2}$$. Find the function $$T(x)$$, where $$x$$ is the distance from the end whose temperature is $$T_{1}$$, and the heat flow density.

A
$$T_{(x)} = 2T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$$.

B
$$T_{(x)} = T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$$.

C
$$T_{(x)} = 3T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$$.

D
$$T_{(x)} = 4T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$$.

Solution