Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 39

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Question 1
1 / -0

If two conducting slabs of thickness $$d_1$$ and $$d_2$$, and thermal conductivity $$K_1$$ and $$K_2$$ are placed together face to face as shown in figure in the steady state temperatures of outer surfaces are $$O_1$$, and $$O_2$$. The temperature of common surface is

A
$$\displaystyle{\frac{K_1O_1d_1 + K_2O_2d_2}{K_1d_1 + K_2d_2}}$$

B
$$\displaystyle{\frac{K_1O_1 + K_2O_2}{K_1 + K_2}}$$

C
$$\displaystyle{\frac{K_1O_1 + K_2O_2}{O_1 + O_2}}$$

D
$$\displaystyle{\frac{K_1O_1d_2 + K_2O_2d_1}{K_1d_1 + K_2d_2}}$$

Solution

In steady state of temperature , the rate of flow of heat in both the slabs will be same .
Let $$\theta$$ be the temperature of common surface ,
then ,
$$Q/t=K_{1}A(O_{1}-\theta)/d_{1}=K_{2}A(\theta-O_{2})/d_{2}$$ ,
or $$\dfrac{K_{1}}{d_{1}}.(O_{1}-\theta)=\dfrac{K_{2}}{d_{2}}.(\theta-O_{2})$$ ,
or $$\dfrac{K_{1}O_{1}}{d_{1}}+\dfrac{K_{2}O_{2}}{d_{2}}=\theta(\dfrac{K_{1}}{d_{1}}+\dfrac{K_{2}}{d_{2}})$$ ,
or $$\theta=\dfrac{K_{1}O_{1}d_{2}+K_{2}O_{2}d_{1}}{K_{1}d_{1}+K_{2}d_{2}}$$
Question 2
1 / -0

If T represents the absolute temperature of an ideal gas, the volume coefficient of thermal expansion at constant pressure is proportional to:

A
T

B
T$$^2$$

C
1/T

D
1/T$$^2$$

Solution

Coefficient of thermal expansion at constant pressure is:

$$\implies { \alpha }_{ V }=\dfrac { 1 }{ V } \dfrac { dV }{ dT } $$

$$ PV=nRT$$

$$ V=\dfrac { nRT }{ P } $$

$$ { \alpha }_{ V }=\dfrac { P }{ nRT } \dfrac { dV }{ dT } $$

$$ { \alpha }_{ V }\rightarrow \dfrac { 1 }{ T } $$
Question 3
1 / -0

Two models of a windowpane are made. In one model, two identical glass panes of thickness 3 mm are separated with an air gap of 3 mm. This composite system is fixed in the window of a room. The other model consists of a single glass pane of thickness 6 mm, the temperature difference being the same as for the first model. The ratio of the heat flow for the double pane to that for the single pane is ($$K_{glass}\, =\, 2.5\, \times\, 10^{- 4}$$ cal / m. $$^{\circ}C$$ and $$K_{air}\, =\, 6.2\, \times\, 10^{- 6}$$ cal/s.m. $$^{\circ} C$$)

A
1/20

B
1/70

C
31/1312

D
31/656

Solution

Thermal Resistance$$R_A= 2R_g+R_{Air}$$
Thermal Resistance$$R_B=\dfrac { 6mm }{ { K }_{ g }A } $$
$$RA=\dfrac { 2\times 3mm }{ { K }_{ g }A } +\dfrac { 3mm }{ { K }_{ air }A }$$
The ratio of heat flow
$$\dfrac { { i }_{ A } }{ { i }_{ B } } =\dfrac { \Delta T\times { R }_{ b } }{ { R }_{ A }\times \Delta T } $$
$$=\dfrac { 6 }{ { K }_{ g }A\times \left( \dfrac { 6 }{ { K }_{ g }A } +\dfrac { 3 }{ { K }_{ air }A } \right) } $$
$$=\dfrac { 2 }{ { K }_{ g }\left( \dfrac { 2 }{ { K }_{ g } } +\dfrac { 1 }{ { K }_{ air } } \right) } $$
$$=\dfrac { 2\times { K }_{ air } }{ { { 2K }_{ air }+K }_{ g } } $$
$$=\dfrac { 2\times 6.2\times { 10 }^{ -6 } }{ 2\times 6.2\times { 10 }^{ -6 }+2.5\times { 10 }^{ -4 } } $$
$$=\dfrac { 2\times 6.2\times { 10 }^{ -2 } }{ 2\times 6.2\times { 10 }^{ -2 }+2.5 } $$
$$=\dfrac { 0.124 }{ 0.124+2.5 } =\dfrac { 0.124 }{ 2.624 } $$
$$=\dfrac { 124 }{ 2624 }=\dfrac { 31 }{ 656 } $$
Question 4
1 / -0

Three copper rods and three steel rods each of length $$l$$ = 10 cm and area of cross - section $$1\, cm^2$$ are connected as shown If ends A and E are maintained at temperatures $$125^{\circ}$$C and $$o^{\circ}$$C respectively, calculate the amount of heat flowing per second from the hot to cold function. $$[K_{cu}\, =\, 400\, W/m-K,\, K_{steel}\, =\, 50\, W/m-K]$$

A
2 watt

B
9 watt

C
4 watt

D
10 watt

Solution

$$R_{steel}\, =\, \displaystyle \frac{L}{KA}\, =\,\frac{10^{-1}m}{50(w/m-^{\circ}C)\, \times\, 10^{-4}m^2}\, =\, \frac{1000}{50}\, ^{\circ} C/W$$
Similarly $$R_cu\, =\,\displaystyle \frac{1000}{50}\, ^{\circ} C/W$$
Junction C and 0 are identical in every respect and both will have same temperature. Consequently.the rod CD is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis.Now rod BC and CE are in series their equivalent resistance is $$R_1 \, =\, R_s\, +\, R_{cu}$$ similarly rods SO and DE are in series with same equivalent resistance $$R_1 \, =\, R_s\, +\, R_{cu}$$ these two are in parallel giving an equivalent resistance of
$$\displaystyle \frac{R_1}{2}\, =\, \frac{R_s\, +\,R_{cu}}{2}$$
This resistance is connected in series with rod AB. Hence the net equivalent of the combination is
$$R\, =\, R_{steel}\, +\, \displaystyle \frac{R_1}{2}\, =\, \frac{3R_{steel}\, +\, R_{cu}}{2}\, =\, 500 \left ( \frac{3}{50}\, +\, \frac{1}{400} \right ) ^{\circ}C/W$$
Now $$i\, =\, \displaystyle \frac{T_H\, -\,T_c}{R}\, =\,\frac{125^{\circ}C}{{500\left ( \frac{3}{50}\, +\,\frac{1}{400} \right )}\,^{\circ}C/W}\, =\,4\, watt$$
Question 5
1 / -0

A manufacture marks the thermometer wrongly. At 0$$^o$$ C it reads -10$$^o$$ C, at 100$$^o$$ C it reads 85$$^o$$C. Then the reading at 50$$^o$$ C will be :

A
40$$^o$$ C

B
32.5$$^o$$ C

C
37.5$$^o$$ C

D
42.5$$^o$$ C

Solution

From the interpolation we can get the value of x.

$$ \dfrac { { 100 }^{ o }-{ 50 }^{ o } }{ { 85 }^{ o }-{ x }^{ o } } =\dfrac { { 50 }^{ o }-{ 0 }^{ o } }{ { x }^{ o }-{ 10 }^{ o } } $$

$$ { x }^{ o }={ 37.5 }^{ o }$$
Question 6
1 / -0

A brass boiler has a base area of $$0.15 m^2$$ and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. The temperature (approximately) of the part of the flame in contact with the boiler is (Thermal conductivity of brass $$= 107.4Js^{-1}m^{-1} {\;} ^oC^{-1}$$, Heat of vaporization of water $$= 2256\times 10^3J kg^{-1})$$.

A
2400 K

B
$$240^oC$$

C
$$2400^oC$$

D
$$24000^oC$$

Solution

Rate of supply of heat of water = Rate at which is conducted by boiler
$$\Rightarrow L=\cfrac { dm }{ dt } =\cfrac { d\theta }{ dt } =\cfrac { kA(\theta -100) }{ t } \\ \Rightarrow \cfrac { 6\times 2256\times { 10 }^{ 3 } }{ 60 } =\cfrac { 107.4\times 0.15\times (\theta -100) }{ { 10 }^{ -2 } } \\ \Rightarrow 225600=\cfrac { 107.4\times 0.15\times (\theta -100) }{ { 10 }^{ -2 } } \\ \Rightarrow \theta -100=140.3\\ \Rightarrow \theta =240℃$$
Question 7
1 / -0

Absolute zero corresponds to :

A
$$- 273 K$$

B
$$- 273^{\circ}C$$

C
$$273^{\circ} F$$

D
None of these

Solution

Hint: Absolute temperature is defined as 0K.
Step 1: $$\textbf{Explanation:}$$

We know that absolute temperature is $$0K$$.

So now we will covert this temperature in Kelvin to degree Celsius, we know that $${{T}_{K}}=273+{{T}_{C}}$$ (Where $${{T}_{K}}$$ is temperature in Kelvin and $${{T}_{C}}$$ is temperature in Celsius.)

$$\Rightarrow 0=273+{{T}_{C}}$$

$$\Rightarrow {{T}_{C}}=-273{{C}^{\circ }}$$
$$\textbf{Thus, option (B) is correct.}$$
Question 8
1 / -0

Which of the following curve represent variation of density of water with temperature best :

A

B

C

D

Solution

$$Answer:-$$ D
This is due to anomalous expansion of water.
The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from $$4^0$$C to $$0^0$$C, and it becomes less dense. The density becomes less and less as it freezes because molecules of water normally form open crystal structures when in solid form.
So, maximum density of water is observed at $$4^0C$$
Question 9
1 / -0

A thermometer uses 'density of water' as thermometric property. The actual reading in the thermometer is 'height of water' (h) which is inversely proportional to density of water (d). In a certain temperature range, density of water varies with temperature as shown. The graph is symmetric about the maximum
Two identical bodies ( of same mass and specific heat ) at different temperatures $$T_1\, and\, T_2$$ show the same reading of height $$h_1\, =\,h_2$$ the thermometer. The bodies are brought into contact and allowed to reach thermal equilibrium. The thermometer reading 'height of water for final equilibrium state $$h_1$$ satisfies.

A
$$h_f\, = \displaystyle \frac{h_1\, +\, h_2}{2}\, =\, h_1\, =\, h_2$$

B
$$h_f\, <\, h_1\, =\, h_2$$

C
$$h_f\, >\, h_1\, =\, h_2$$

D
$$h_f$$ may be greater or less than $$h_1\, =\, h_2$$, depending on the specif heat of the bodies

E
Information is not enough.

Solution
Question 10
1 / -0

The boiling point of mercury is 367$$^o$$C. A mercury thermometer can be used to measure a temperature of 500$$^o$$C;

A
by filling the space above mercury with oxygen at high pressure

B
by filling the space above mercury with nitrogen at low pressure

C
by filling the space above mercury with nitrogen at high pressure

D
by keeping the space above mercury as vacuum

Solution

The mercury thermometer can be used above its boiling point by filling up the space above mercury with nitrogen at high pressure because as pressure increases, boiling point also increases.