Thermal Properties of Matter Test

Result

Thermal Properties of Matter Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two rods of same length and cross sections are joined end to end. Their thermal conductivities are in the ratio $$2 : 3$$. If the free end of the first rod is at $$0^{o}$$C and free end of the second rod is at $$100^{o}$$C , the temperature at the junction of the two rods after attaining steady state is:

A
$$33.33^{o}C$$

B
$$40^{o}C$$

C
$$50^{o}C$$

D
$$60^{o}C$$

Solution

From the Fourier's law we know that heat transfer through a plane wall is given as:

$${Q}=\dfrac{KA{\Delta}T}{l}$$

Here length and area both are same for both the rod.

So, $${Q}={{K}_{1}{\Delta}T}$$

So, $${Q}_{1}={{K}_{1}({T}_{1}-{T})}$$

$${Q}_{2}={{K}_{2}({T}-{T}_{2})}$$

On equating both $${Q}_{1}$$ and $${Q}_{2}$$ we get

$${{K}_{1}({T}_{1}-{T})}={{K}_{2}({T}-{T}_{2})}$$

$$\dfrac{{K}_{1}}{{K}_{2}}=\dfrac{{T}-{T}_{2}}{{T}_{1}-{T}}$$

$$\dfrac{2}{3}=\dfrac{{T}-{100}}{{0}-{T}}$$

$${T}={60}^{0}C$$
Question 2
1 / -0

Match List I and List II
List-I List-II
a) P-V graph(T is constant) e) St. line cutting temp axis at - 273$$^{0}$$
b) P-T graph(V is constant) f) Rectangular hyperbola
c) V-T graph(pressure P is constant) g) A st.line parallel to axis
d) PV- P garph h) St. line passing trhough origin (T is constant)

A
a-g,b-e,c-h, d-f.

B
a-h, b-f c-g, d-e.

C
a-e, b-g,c-f, d-h.

D
a-f, b-h, c-e,d-g.

Solution

(a) P-V graph (T is constant):
$$PV = nRT$$
at constant temperature
$$PV = constant$$
$$P \alpha \frac{1}{V}$$
so it will be Rectangular hyperbola.

(b) P-T Graph (V is constant) :
$$PV = nRT$$
at constant volume
$$P \alpha T$$
pressure is proportional to T, so it will be straight line passing through origin.

c) V-T graph(P is constant) :
$$PV = nRT$$
at constant pressure
$$V \alpha T$$
volume is proportional to T(Kelvin),so it will be straight line passing through origin, but if temperature is taken in $$^{o}C$$, then it will cut temperature axis at $$-273 ^{o}C$$

(d) PV - P Graph (T is constant) :
$$PV = nRT$$
at constant Temperature
$$PV = constant$$
$$PV$$ is constant for constant temperature,so it will be straight line parallel to P axis.
Question 3
1 / -0

The thermal radiations are similar to :

A
x-rays

B
cathode rays

C
$$\alpha $$- rays

D
$$\gamma $$- rays

Solution

Among the above four options: X-ray can only act as heat rays. $$\alpha$$-rays are nuclear emissions, cathode rays are simply rays made up of electrons. $$\gamma$$-rays are simply photon rays, $$\gamma$$-rays have very high penetrating power so it cannot act as heat ray, So X-ray is the appropriate answer.
Question 4
1 / -0

If we place our hand below a lighted electric bulb. We feel warmer because of :

A
convection

B
radiation

C
conduction

D
both A and B
Solution
There are three ways by which heat transfers:
Conduction: for which the two bodies need to be in contact

Convection: for which there needs to be a material medium between the two bodies

Radiation: which occurs if there is a temperature difference, irrespective of whether there is a material medium or not.

In this case, there is a medium between the two. But the two bodies are not in contact.
Hence, there is radiation and convection which matches with option D.
Question 5
1 / -0

Two metal plates of same area and ratio of thickness $$1 : 2$$ and having ratio of thermal conductivities $$1 : 2$$ are in contact. If the free sides of the first plate is maintained at $$-20^{o}$$C and the free side of the other plate is maintained at $$+40^{o}$$C ,the temperature of the common surface after attaining steady state is:

A
$$0^{o}C$$

B
$$-10^{o}C$$

C
$$10^{o}C$$

D
$$20^{o}C$$

Solution

From the fourier's law we know that heat transfer through a plane wall is given as:
$${Q}=\dfrac{KA{\Delta}T}{l}$$
Here, area are same for both the rod,
$${Q}=\dfrac{{{K}4{\Delta}T}}{l}$$
$${Q}_{1}=\dfrac{{K}_{1}({T}_{1}-{T})}{{l}_{1}}$$
$${Q}_{2}=\dfrac{{K}_{2}({T}-{T}_{2})}{{l}_{2}}$$

On equating both $${Q}_{1}$$ and $${Q}_{2}$$ we get
$$\dfrac{{K}_{1}({T}_{1}-{T})}{{l}_{1}}=\dfrac{{K}_{2}({T}-{T}_{2})}{{l}_{2}}$$
$$({T}_{1}-{T})=({T}-{T}_{2})$$
$${-20}-{T}={T}-{40}$$
$${T}={10}^{\circ}C$$
Question 6
1 / -0

Three rods made of same material and having the same cross section and lengths have been joined as shown in the figure. The temperature at the junction of the rods will be :

A
$$45^{o}C$$

B
$$60^{o}C$$

C
$$30^{o}C$$

D
$$20^{o}C$$

Solution

Let the temperature of the junction be $${\theta}$$
So assume that total incoming heat at the junction will be equal to total outgoing heat.
So, $$({T}_{1}-{\theta})=({\theta}-{T}_{2})+({\theta}-{T}_{3})$$
$$({0}-{\theta})=({\theta}-{90})+({\theta-90})$$
$${\theta}={60}^{0}C$$
$$\dfrac{KA(\theta - 0)}{\ell} = \dfrac{KA(90-\theta)}{\ell} = \dfrac{KA(90-\theta)}{\ell}$$
Question 7
1 / -0

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 constant temperatures of 270$$^{o}$$C and 0$$^{o}$$C respectively.The thermal conductivity of glass is 0.8 and of air 0.08Wm$$^{-1}$$K$$^{-1}$$.

The rate of flow of heat through the window pane is nearly equal to:

A
$$1000Js^{-1}$$

B
$$1224Js^{-1}$$

C
$$3000Js^{-1}$$

D
$$4000Js^{-1}$$

Solution

Given: 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 constant temperatures of $$270^oC$$ and $$0^oC$$ respectively.The thermal conductivity of glass is 0.8 and of air $$0.08Wm^{−1}K^{−1}$$
To find the rate of flow of heat through the window pane.
Solution:
As per the given criteria,
Area of the each glass sheet, $$A=1m^2$$
Thermal conductivity of air, $$k_a=0.08Wm^{−1}K^{−1}$$
Thermal conductivity of glass, $$k_g=0.8Wm^{−1}K^{−1}$$
Temperature at room-glass interface, $$T_1=270^\circ C +273=543K$$
Temperature at glass-outdoor interface, $$T_4=0^\circ C+273=273K$$
Thickness of the glass sheet, $$d_g=0.01m$$
Thickness of the air space, $$d_a=0.05m$$
By conduction, the equation of heat flow are:
$$\dfrac {dQ_1}{dt}=\dfrac {k_gA(T_1-T_2)}{d_g}$$
By substituting the values, we get
$$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-T_2)}{0.01}......(i)$$
Similarly at second interface, we get
$$\dfrac {dQ_2}{dt}=\dfrac {k_aA(T_2-T_3)}{d_a}$$
By substituting the values, we get
$$\dfrac {dQ_2}{dt}=\dfrac {0.08\times1\times(T_2-T_3)}{0.05}......(ii)$$
Similarly at third interface, we get
$$\dfrac {dQ_3}{dt}=\dfrac {k_gA(T_3-T_4)}{d_g}$$
By substituting the values, we get
$$\dfrac {dQ_3}{dt}=\dfrac {0.8\times1\times(T_3-273)}{0.01}......(iii)$$
But in steady state,
$$\dfrac {dQ_1}{dt}=\dfrac {dQ_2}{dt}=\dfrac {dQ_3}{dt}$$
So equating eqn(i) and (ii), we get
$$\dfrac {0.8\times1\times(543-T_2)}{0.01}=\dfrac {0.08\times1\times(T_2-T_3)}{0.05}\\\implies 0.05(0.8\times(543-T_2))=0.01(0.08\times(T_2-T_3))\\\implies 0.04(543-T_2)=0.0008(T_2-T_3)\\\implies (543-T_2)=\dfrac {0.0008}{0.04}(T_2-T_3)\\\implies 543-T_2=0.02T_2-0.02T_3\\\implies 1.02T_2-0.02T_3=543......(iv)$$
Now equating eqn(ii) and (iii), we get
$$\dfrac {0.08\times1\times(T_2-T_3)}{0.05}=\dfrac {0.8\times1\times(T_3-273)}{0.01}\\\implies 0.01(0.08\times(T_2-T_3))=0.05(0.8\times(T_3-273)\\\implies 0.0008(T_2-T_3)=0.04(T_3-273)\\\implies\dfrac {0.0008}{0.04}(T_2-T_3)= (T_3-273)\\\implies0.02T_2-0.02T_3=T_3-273\\\implies 0.02T_2-1.02T_3=273......(v)$$
Multiply, eqn(iv) with $$0.02$$ and eqn(v) with $$1.02$$ and by subtracting them,we get
$$0.0204T_2-0.0004T_3=10.86$$
$$\underline{-(0.0204T_1-1.0404T_3=278.46)}\\\implies 1.04T_3=-267.6\\\implies T_3=-257.3K$$
Substituting the value of $$T_3$$ in eqn(v), we get
$$0.02T_2-1.02(-257.3)=273\\\implies 0.02T_2=273-262.446\\\implies T_2=527.7K$$
So the rate of flow of the heat through the window pane is given by,
$$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-T_2)}{0.01}$$
Substituting the value of $$T_2$$, we get
$$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-527.7)}{0.01}\\\implies \dfrac {dQ_1}{dt}=1224W$$
is the required rate of flow of heat.
Question 8
1 / -0

Two rods (one semi-circular and other straight) of same material and of same cross sectional area are joined as shown in the figure. The points $$A$$ and $$B$$ are maintained at different temperatures. The ratio of heat transferred through a cross-section of a semicircular rod to the heat transferred through a cross section of the straight rod in a given time is

A
$$2:$$$$\pi $$

B
$$1:2$$

C
$$\pi $$$$:2$$

D
$$3:2$$

Solution

We know that for semi-circular rod amount of heat generated is given as:

$${ Q }_{ 1 }=\dfrac { KA({ T }_{ 1 }-{ T }_{ 2 })t }{ \pi R } $$

For straight rod amount of heat generated is given as:

$${ Q }_{ 2 }=\dfrac { KA({ T }_{ 1 }-{ T }_{ 2 })t }{ 2R }$$

So taking there ratio, we find $$\dfrac { { Q }_{ 1 } }{ { Q }_{ 2 } } =\dfrac { 2 }{ \pi } $$
Question 9
1 / -0

Statement-1
Radiation involves transfer of heat by electromagnetic waves.
Statement-2
Electromagnetic waves do not required any material medium for propagation.

A
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1.

B
Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Staement-1.

C
Statement-1 is true, Statement-2 is false.

D
Statement-1 is false, Statement-2 is true.

Solution

Electromagnetic waves are formed due to simultaneous oscillations in electric and magnetic field (both are perpendicular to each other and perpendicular to the propagation of the em wave). Both electric and magnetic fields can exist in a vacuum and do not need a material medium to exist or oscillate. So an electromagnetic wave can propagate without a medium.
These radiations carry energy with them, so when a body releases radiation it loses energy, which manifests into a temperature loss. When radiation (energy) is absorbed, the body gains energy which is manifested in the form of a temperature gain. This is basically a heat transfer with radiation.
Both statement 1 and Statement 2 are true, but Statement-2 is not the correct explanation for Staement-1.
Option B is correct.
Question 10
1 / -0

Three rods of same material and having same crosssection have been joined as shown in fig.Each rod is of same length. The left and right rods are kept at 0$$^{o}$$C and 90$$^{o}$$C respectively.The temperature of the junction of the three rods will be

A
$$45^{o}C$$

B
$$60^{o}C$$

C
$$30^{o}C$$

D
$$20^{o}C$$

Solution

Let the temperature of the junction will be $${X}$$
So assume that total incoming temperature at the junction will be equal to total outgoing temperature.
So $$({T}_{1}-{X})=({X}-{T}_{2})+({X}-{T}_{3})$$
$$({0}-{X})=({X}-{90})+({X-90})$$
$${X}={60}^{0}C$$