Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 48

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Question 1
1 / -0

Three metal rods made of copper ,aluminium and brass, each 20 cm long and 4 cm in diameter,are placed end to end with aluminium between the other two.The free ends of copper and brass are maintained at 100 and $$0^0$$ C respectively. Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. The equilibrium temperatures of the copper aluminium and aluminium brass junctions are respectively.

A
$$68^0$$ C and $$75^0$$

B
$$75^0$$C and $$68^0$$

C
$$57^0$$C and $$86^0$$

D
$$86^0$$C and $$57^0$$

Solution
Question 2
1 / -0

By the ideal gas law, the pressure of $$0.60$$ moles $${NH}_{3}$$ gas in a $$3.00\ L$$ vessel at $${25}^{o}C$$ is, given that $$R=0.082\ L$$ atm $${mol}^{-1}{k}^{-1}$$:

A
$$48.9\ atm$$

B
$$4.89\ atm$$

C
$$0.489\ atm$$

D
$$489\ atm$$

Solution

Pressure is given as $$P=\dfrac{nRT}{V}=\dfrac{0.6\times .082\times (25+273)}{3}=4.89atm$$
Option B is correct.
Question 3
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One kg of a diatomic gas is at a pressure of $$8\times {10}^{4}N/{m}^{2}$$. The density of gas is $$4kg/{m}^{2}$$. What is the energy of the gas due to its thermal motion?

A
$$6\times {10}^{4}J$$

B
$$7\times {10}^{4}J$$

C
$$3\times {10}^{4}J$$

D
$$5\times {10}^{4}J$$

Solution
Question 4
1 / -0

Time taken by a $$836$$ W heater to heat one litre of water from $$10^o$$C to $$40^o$$C is?

A
$$50$$s

B
$$100$$s

C
$$150$$s

D
$$200$$s

Solution

According to the given question we should find the time taken
It is given that : $$\Rightarrow P = 836$$W
$$\Rightarrow$$ mass of water $$= 1L = 1000$$g
$$\Rightarrow$$ temperature rise $$=\left({\theta}_{1}-{\theta}_{2}\right) =\left(40-10\right) ={30}{\circ}$$
$$\Rightarrow$$ specific heat of water is $$C = 1$$ cal$${g}^{-1}{c}^{-1}$$
Now,heat produced by the heater in time $$t$$ is $$= P \times T = 836 t$$ joules ( we should convert to calories )
$$=\dfrac{836 t}{4.2}$$ ......$$(1)$$ calories---------------------(1)
Now, heat taken by the water $$= mc\left({\theta}_{1}-{\theta}_{2}\right)= 1000 \times 1 \times 30=30000$$-----------$$(2)$$
we know that ,heat produced by the heater in time $$t =$$ heat taken by the water implies,$$\dfrac{836 t}{4.2} = 30000$$
$$\Rightarrow t = \dfrac{\left(30000 \times 4.2\right)}{836}$$
$$t =\dfrac{126000}{836}$$
$$t = 150.7$$ seconds .
$$\therefore$$ the time taken is $$150.7$$ seconds.
Question 5
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A spherical half of volume $$1000\ cm^3$$ is subjected to a pressure of $$10$$ atmosphere. The change in volume is $$10^{-3}\ cm^{-3}$$. If the ball is made of iron find its bulk modulus.(Atmospheric pressure $$=1\times 10^{3}\ Nm^{-2})$$

A
$$3\times 10^{11}\ N/m^2$$

B
$$2\times 10^{11}\ N/m^2$$

C
$$1\times 10^{11}\ N/m^2$$

D
$$4\times 10^{11}\ N/m^2$$

Solution

$$\triangle V=10^{-2}cm^3$$
$$V=1000cm^3$$
$$P=10atm$$
$$=10\times 10^5Nm^{-2}$$
$$B =\cfrac{P}{\triangle V/V}$$
$$=\cfrac{10\times 10^5\times 1000}{10^{-2}}$$
$$B=10^{11}Nm^{-2}$$
Question 6
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Three perfect gasses at absolute temperatures $$T_1$$, $$T_2$$ and $$T_3$$ are mixed. If number of molecules of the gasses are $$n_1$$, $$n_2$$ and $$n_3$$ respectively then temperature of mixture will be (assume no loss of energy)

A
$$\dfrac{ T_{1}+ T_{2}+T_{3}}{3}$$

B
$$\dfrac{ { n }_{ 1 }^{ 2 } T_{1}+ { n }_{ 2 }^{ 2 } T_{2}+ { n }_{ 3 }^{ 2 } T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

C
$$\dfrac{ { n }_{ 1 } T_{1}+ { n }_{ 2 }T_{2}+ { n }_{ 3 }T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

D
$$\dfrac{ T_{1}+ T_{2}+T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

Solution

For perfect gas,
Kinetic Energy of n molecule, $$K.E=n\left( \dfrac{1}{2}{{K}_{B}}T \right)$$
Where, $${{K}_{B}}$$ is Boltzmann constant
If there is no loss of energy.
Total kinetic energy of mixture is sum of each gas kinetic energy.
$$ {{n}_{total}}K.{{E}_{total}}={{n}_{1}}K.{{E}_{1}}+{{n}_{2}}K.{{E}_{2}}+{{n}_{3}}K.{{E}_{3}} $$
$$ \left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\left( \dfrac{1}{2}{{K}_{B}}T \right)={{n}_{1}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{1}} \right)+{{n}_{2}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{2}} \right)+{{n}_{3}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{3}} \right) $$

$$ T=\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}} $$
Question 7
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Two diagonally opposite comers of a square made of four thin rods of the same material, same dimensions are at temperature$$ 40^o C$$ and $$10^o C$$. If only heat conduction takes place, then the temperature difference between the other two corners will be:

A
$$0^{o}C$$

B
$$10^{o}C$$

C
$$25^{o}C$$

D
$$15^{o}C$$

Solution

energy flow rate is $$\dfrac{dQ}{dt}=\dfrac{KA(T_{1}-T_{2})}{L}$$
let T be the temperature of the other diagonal of the square.
The rate of flow of heat from one rode is equal to another rod.
$$\dfrac{KA(40^0-T)}{L}=\dfrac{KA(T-10^0)}{L}$$
$$2T=30$$
$$T=15^0$$
hence D is the correct option.
Question 8
1 / -0

In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?

A

$$1.12\times 10^4 J$$

B

$$1.40\times 10^4$$

C
$$2.12\times 10^4$$J

D
$$3.12×10^4 J$$

Solution

Since oxygen $$(O_2)$$ is a diatomic gas
$$\therefore C_v=5/2R$$
So, $$C_p=C_v+R\\ \quad=5/2R+R\\ \quad =7/2R$$
At constant pressure
If we double the volume the temperature will be doubled.
$$T_i=11.0°C=(11.0+273)K=284K$$
So, $$T_p=2(284)K=568K$$
To obtain heat, $$Q=nC_p(T_p-T_i)$$
$$Q=(1.35mol)(7/2R)(508K-284K)\\Q=1.12\times10^4J$$
Question 9
1 / -0

A vessel has $$6g$$ of oxegen at pressure $$P$$ and temperature $$400\ K$$. A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is $$P/2$$ and temperature is $$300\ K$$

A
$$3g$$

B
$$2g$$

C
$$4g$$

D
$$5g$$

Solution

From ideas gas equation
$$ PV=nRT $$
$$ PV=\dfrac{m}{{{M}_{o}}}RT $$
$$ m=\dfrac{PV{{M}_{o}}}{RT}\ $$
In first event
$$ {{m}_{1}}=\dfrac{{{P}_{1}}V{{M}_{o}}}{R{{T}_{1}}} $$
$$ 6\ gram=\dfrac{PV{{M}_{o}}}{R\times 400}\ $$
$$ \dfrac{{{\operatorname{PVM}}_{o}}}{R}=400\times 6\ .\ .....\ (1) $$
In the second event when hole is made:
$$ {{m}_{2}}=\dfrac{{{P}_{2}}V{{M}_{o}}}{R{{T}_{2}}} $$
$$ {{m}_{2}}=\dfrac{\dfrac{P}{2}V{{M}_{o}}}{R\times 300}\ =\dfrac{1}{300\times 2}\dfrac{PV{{M}_{o}}}{R} $$
$$ {{m}_{2}}=\dfrac{1}{300\times 2}\times \left( 400\times 6 \right)=4\ gram $$

Leak out oxygen is $$\left( 6gram-4\,gram \right)\ =\ 2\,gram$$
Question 10
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If a thermometer reads freezing point of water as $$20^{\circ}C$$ and boiling point at $$150^{\circ}C$$ how much thermometer read when the actual temperature is $$60^{\circ}C$$.

A
$$98^{\circ}C$$

B
$$110^{\circ}C$$

C
$$40^{\circ}C$$

D
$$60^{\circ}C$$

Solution

We let the unknown degree scale be called $${x}^{\circ}$$
Now we know that the freezing point$$={20}^{\circ}x$$
The boiling point$$={150}^{\circ}x$$
This also means that $${20}^{\circ}x= {0}^{\circ}$$ C($$\because$$ on a Celcius Scale, the freezing point is $${0}^{\circ}$$)
And $${150}^{\circ}x={100}^{\circ}$$C (On a Celcius Scale, boiling point is $${100}^{\circ}$$)
Also $${(150-20)}^{\circ}x= {100}^{\circ}$$C ($$\because$$ on a Celcius Scale $${100}^{\circ}-{0}^{\circ}={100}^{\circ}$$)
So, if $${100}^{\circ}$$C$$={130}^{\circ}x$$
Then $${1}^{\circ}$$C$$={1.3}^{\circ}x$$
To go from $$C$$ degree to $$x$$, we need to use:
$$x=1.3C+20$$
And $$C=\dfrac{(x-20)}{1.3}$$
So if the temperature given is $${60}^{\circ}$$C, then we calculate $$x$$ as:
$$C=\dfrac{(x-20)}{1.3}$$
$$\Rightarrow 60=\dfrac{(X-20)}{1.3}$$
$$\Rightarrow 60\times 1.3=x-20$$
$$\Rightarrow 78=x-20$$
$$\Rightarrow x= {98}^{\circ}$$
So $${60}^{\circ}$$C will be equal to $${98}^{\circ}$$

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Thermal Properties of Matter Test - 48

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Thermal Properties of Matter Test - 48

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Question 1

$$68^0$$ C and $$75^0$$

$$75^0$$C and $$68^0$$

$$57^0$$C and $$86^0$$

$$86^0$$C and $$57^0$$

Solution

Question 2

$$48.9\ atm$$

$$4.89\ atm$$

$$0.489\ atm$$

$$489\ atm$$

Solution

Question 3

$$6\times {10}^{4}J$$

$$7\times {10}^{4}J$$

$$3\times {10}^{4}J$$

$$5\times {10}^{4}J$$

Solution

Question 4

$$50$$s

$$100$$s

$$150$$s

$$200$$s

Solution

Question 5

$$3\times 10^{11}\ N/m^2$$

$$2\times 10^{11}\ N/m^2$$

$$1\times 10^{11}\ N/m^2$$

$$4\times 10^{11}\ N/m^2$$

Solution

Question 6

$$\dfrac{ T_{1}+ T_{2}+T_{3}}{3}$$

$$\dfrac{ { n }_{ 1 }^{ 2 } T_{1}+ { n }_{ 2 }^{ 2 } T_{2}+ { n }_{ 3 }^{ 2 } T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

$$\dfrac{ { n }_{ 1 } T_{1}+ { n }_{ 2 }T_{2}+ { n }_{ 3 }T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

$$\dfrac{ T_{1}+ T_{2}+T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$$

Solution

Question 7

$$0^{o}C$$

$$10^{o}C$$

$$25^{o}C$$

$$15^{o}C$$

Solution

Question 8

$$1.12\times 10^4 J$$

$$1.40\times 10^4$$

$$2.12\times 10^4$$J

$$3.12×10^4 J$$

Solution

Question 9

$$3g$$

$$2g$$

$$4g$$

$$5g$$

Solution

Question 10

$$98^{\circ}C$$

$$110^{\circ}C$$

$$40^{\circ}C$$

$$60^{\circ}C$$

Solution

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