Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 74

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Question 1
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The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. If the atmosphere pressure is $$Hm$$ of mercury and the density of mercury is $$n$$ times that of lake water the depth of lake is :

A
$$H/n$$

B
$$nH/2$$

C
$$nH$$

D
$$2nH$$

Solution
Question 2
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Directions For Questions

In a container of negligible heat capacity, $$200\ gm$$ ice at $$0^{o}C$$ and $$100\ gm$$ steam at $$100^{o}C$$ are added to $$200\ cm$$ of water that has temperature $$55^{o}C$$. Assume no heat is lost to the surrounding and the pressure in the container is constant $$1.0\ atm$$.
$$(L_{f}=80\ cal/gm, L_{v}=540\ cal/gm, S_{w}=1\ cal/gm\ ^{o}C)$$

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What is the final temperature of the system ?

A
$$48^{o}C$$

B
$$72^{o}C$$

C
$$94^{o}C$$

D
$$100^{o}C$$

Solution
Question 3
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Directions For Questions

A $$500\ g$$ teapot and an insulated thermos are in a $$20^{o}C$$ room. The teapot is filled with $$1000\ g$$ of the boiling water. $$12$$ tea bags are then placed into the teapot. The brewed tea is allowed to cool to $$80^{o}C$$, then $$250\ g$$ of the tea is poured from the teapot into the thermos. The teapot us then kept on an insulated warmer that transfers $$500\ cal/min$$ to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that the water, and a negligible effect on the temperature. The specific heat of teapot is $$0.17\ J/g\ K$$ and that of water is $$4.18\ J/g\ K$$. The entire procedure is done under atmosphere pressure. There are $$4.18\ J$$ in one caloric.

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If, after some of the tea has been transferred to the thermos (as described in the passage), the teapot with its contents (at a temperature of $$80^{o}C$$) was placed or the insulated warmer for $$5$$ minutes, what would be the temperature at the end of this $$5$$ minutes period (Assume that no significant heat transfer occurs with the surrounding):

A
$$80.7\ ^{o}C$$

B
$$82.5\ ^{o}C$$

C
$$83.2\ ^{o}C$$

D
$$95.2\ ^{o}C$$

Solution
Question 4
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The only possibility of heat flow in a thermos flask is through its cork which is $$75 cm$$$$^2$$ in the area and $$5 cm$$ thick. Its thermal conductivity is $$0.0075$$ cal/cm-s-$$^o$$C. The outside temperature is $$40^o$$C and the latent heat of ice is $$80 cal/g$$. Time is taken by $$500 g$$ of ice at $$0^o$$C in the flask to melt into the water at $$0^o$$C is

A
$$2.47 h$$

B
$$4.27 h$$

C
$$7.42 h$$

D
$$4.72 h$$

Solution

Given,
$$(m=500g,L=80 cal/g,k=0.0075cal/cm-s^0C,A=75cm^{2},\Delta \theta=40^{0}C,\Delta x=5cm,s=?)$$

$$mL = \dfrac{KA \Delta \theta t}{\Delta x}$$

$$\Rightarrow 500 \times 80 = \dfrac{0.0075 \times 75 \times (40 -0)t}{5}$$

$$\Rightarrow t = 8.9 \times 10^3$$

$$ s = 2.47 h$$
Question 5
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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 $$\sqrt 2 T$$ respectively in the steady state. Assuming that only heat conduction takes place temperature of point $$C$$ will be:

A
$$\dfrac {3T}{\sqrt 2 +1}$$

B
$$\dfrac {T}{\sqrt 2 +1}$$

C
$$\dfrac {T}{\sqrt 3 (\sqrt 2 +1)}$$

D
$$\dfrac {T}{\sqrt 2 -1}$$

Solution
Question 6
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A solid aluminium sphere and a solid lead sphere of the same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium = $$
900\space J/kg^{0}C$$ and that of the lead = $$ 130 \space J/kg^{0}C$$. The density of lead = $$10^{4} \space kg/m^{3}$$ and that of aluminum = $$ 2.7 \times 10^{3} \space kg/m^{3}
\space kg/m^{3}$$. Assume that the emissivity of both the
spheres is the same
The ration of the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the lead sphere is

A
$$1000:39$$

B
$$39:1000$$

C
$$1:1$$

D
$$13:39$$

Solution

Let $$\dfrac{d\Delta_1}{dt}$$ be the rate of fall of temperature of an aluminium sphere and $$\dfrac{d\Delta_2}{dt}$$ be the rate of fall of temperature of a lead sphere.
$$ \dfrac {P_1} {p_2} = \dfrac {m_1 S_1 \dfrac{d\theta_1} {dt}} {m_2 S_2 \dfrac{d\theta_2} {dt}} = \dfrac {v_1d_1s_1 \dfrac{d\theta_1} {dt}} {v_2d_2s_2 \dfrac{d\theta_2} {dt}} = 1$$
$$\dfrac {\dfrac {d \theta_1} {dt}} {\dfrac {d \theta_2} {dt}} =\dfrac {S_2} {S_1} \dfrac{d_2} {d_1} = \dfrac{130 \times 2.7} {900 \times 10} = \dfrac {39} {1000}$$
$$[ \therefore V_1 = V_2]$$
Question 7
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Two identical calorimeters, each of water equivalent 100 g and volume 200 cm$$^3$$, are filled with water and a liquid. They are placed in identical constant-temperature enclosures to cool down. The temperatures are plotted at different times (the choice of units are completely arbitrary) as shown in the figure. If the density of the liquid is 800 kgm$$^{-3}$$, then its specific heat capacity is

A
$$8400 J/kg^o$$C

B
$$2100 J/kg^o$$C

C
$$1312.5 J/kg^o$$C

D
$$1680.5 J/kg^o$$C

Solution

From the graph for the same temperature drop, ($$\Delta T$$ say), the respective time taken by the liquid and water are 1 and 2 units, respectively. Average rate of heat losses for the two containers should be the same.
$$\therefore (100 g \times 4200 J/kg^oC + 160 g \times s) (\Delta T/1)$$

$$ = (100 g + 200 g) 4200 J/kg^oC (\Delta T/2)$$

$$\Rightarrow s = 1312.5 J/kg^o$$C
Question 8
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Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at $$100^{0}C$$ and $$0^{0}C$$, respectively. The system is allowed to reach the steady-state condition. Assume there is no loss of heat anywhere.
When a steady-state condition is reached everywhere, which of the following statements is true?

A
No heat is transmitted across the copper-aluminium or aluminium-brass junctions.

B
More heat is transmitted across the copper-aluminium junction than across the aluminium-brass junction.

C
More heat is transmitted across the aluminium-brass junction than the copper-aluminium junction.

D
Equal amount of heat is transmitted at the copper-aluminium and aluminium-brass junctions.

Solution

Under steady-state conditions, the temperatures at all sections in the system remain constant and maintain a constant temperature gradient for a given material. The temperature gradient in copper, aluminium and brass will not be same however, the rate of heat conducted across all sections whether in copper or aluminium or brass will be the same.
Question 9
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A solid aluminium sphere and a solid lead sphere of the same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium = $$
900\space J/kg^{0}C$$ and that of the lead = $$ 130 \space J/kg^{0}C$$. The density of lead = $$10^{4} \space kg/m^{3}$$ and that of aluminum = $$ 2.7 \times 10^{3} \space kg/m^{3}\space kg/m^{3}$$. Assume that the emissivity of both the
spheres is the same
When the temperature of spheres T is not too different from the surrounding temperature, the radiating object obeys

A
Stefan-Boltzmann law

B
Newton's law of cooling

C
Dulong-Petit law

D
Planck's law

Solution

Power radiated by an object to the surrounding temperature $$T_0$$
$$ P = e \sigma A(T^{4} - T^{4}_0)$$
$$ = \sigma eA(T^{2} + T^{2}_0) (T_0 + T_0) (T - T_0)$$
$$ = 4e\sigma \space A \space T^{3}_0 \Delta T$$
$$ P \space \alpha \space \Delta T$$
The net power radiated is approximately proportional to the temperature difference, in agreement with Newton's law of cooling it $$(T - T_0)$$ is small.
Question 10
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An immersion heater, in an insulated vessel of negligible heat capacity, brings 100 g of water to the boiling point from $$16^{0}C$$ in 7 min. Then
Power of heater is nearly

A
$$ 8.4 \times 10^{3}$$

B
$$84 W$$

C
$$ 8.4 \times 10^{3}$$ cal/s

D
$$20 W$$

Solution

In $$7 min$$, temperature of $$100 g$$ of water is raised by $$ (1000 - 16^{0}C) = 84^{0}C$$. The amount of the heat provided by heater

$$ Q_W = C_Wm_W\Delta T = (1 cal/g^{0}C)(100 g)(84^{0}C)$$

$$ = 8.4 \times 10^{3} cal = (8.4 \times 10^{3} \times 4.186) J$$

$$ \simeq 3.5 \times 10^{4} J$$

Power of heater = $$\frac {Q_W} {t_1}$$

$$ = \dfrac {8.4 \times 10^{3} cal} {(7 \times 60)_s} = 20 cal/s$$

$$ = (20 \times 4.18) J/s = 83.6 W = 84 W$$