Heat Test

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Heat Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The door of a running refrigerator inside a closed room was left open. Which of the following observations will be seen?

A
The room will be warmed up gradually.

B
The temperature of the room will remain unaffected.

C
The room will become as cold as the refrigerator.

D
The room will be cooled slightly.

Solution

The room around a refrigerator is warmed as it receives the heat removed from inside the box. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.
Question 2
1 / -0

At what temperature do both the centigrade and fahrenheit thermometers show the same reading?

A
-40^oC

B
40^oC

C
0^oC

D
32^oC

E
-32^oC

Solution

Formula for Fahrenheit, F = 9/5 C + 32
Formula for Centigrade, C =5/9 (F-32)
Let F = C or C= F
F= C= 5/9(F-32)= (5F-160)/9
9F= 5F - 160
9F - 5F = -160
4F = -160
F = -160/4
F= -40 degree answer in Fahrenheit
C= F= 9/5 C + 32
C = (9C + 160)/5
5C = 9C + 160
5C - 9C = 160
- 4C= 160
C=160/-4
C= -40 degree answer in Centigrade
Conclusion: The temperature wherein degrees centigrade is the same in degrees Fahrenheit is - 40 degrees.
Question 3
1 / -0

A liquid coefficient of volume expansion γ is filled in a container of a material having the coefficient of linear expansion α. If the liquid overflows on heating, then

A
γ > 3α

B
γ < 3α

C
γ = 3α

D
γ - 3α

E
Insufficient information

Solution

Volumetric coefficient of expansion of material is 3α
As the system is heated liquid started flowing out of box,it means volumetric coefficient of expansion of liquid is greater than of the container.
Hence γ > 3α.
Question 4
1 / -0

A glass flask is filled up to a mark with 50 ccof mercury at $18^{\circ} \mathrm{C}$. If the flask and content are heated to $38^{\circ} \mathrm{C}$, how much mercury will be above the mark? ( $\alpha$ for glass is $9 \times 10^{-6} / C$ and coefficient of real expansion of mercury is $180 \times 10^{-6} / C$

A
$0.85 c c$

B
$0.46 c c$

C
$0.153 c c$

D
$0.05 c c$

E
None of these

Solution

Volume expansion coefficient of glass $=3 \times$ linear $=3 \alpha$
Change in temperature is $38-18=20^{0} C$
Increment of volume of flask is $\Delta V=50 c c \times 3 \alpha \times 20^{0} C=0.027 c c$
Expansion in mercury $50 c c \times \gamma \Delta t=50 \times 180 \times 10^{-6} \times 20=0.180 \mathrm{cc}$
Apparent increment = 0.180 - 0.027 $=0.153 \mathrm{cc}$
Option $\mathrm{C}$ is correct.
Question 5
1 / -0

Which of the following conditions enables heat to flow from one body to another identical body?

A
The difference in the temperatures of the two bodies

B
The difference in the pressures exerted on the two bodies

C
The difference in the densities of the two bodies

D
The difference in the sizes of the two bodies

E
None of these

Solution

When two identical bodies differ in their temperatures are in contact, heat flows from a body at a higher temperature to the other body at lower temperature.
Question 6
1 / -0

The coefficient of volume expansion of mercury is 20 times the coefficient of linear expansion of glass. If the volume of mercury that must be poured into a glass vessel of volume V so that the volume above mercury may remain constant at all temperature is $\frac{x V}{20}$ . Find x.

A
3

B
6

C
9

D
12

E
15

Solution

$\gamma_{\text {mercury }}=20 \alpha_{\text {glass }}=\frac{20}{3} \gamma_{\text {glass }}$
Let the volume of mercury is $V_{\text {mercury}}$
Since the volume above mercury remains same,
$\gamma_{\text {mercury }} V_{\text {mercury }}=\gamma_{\text {glass }} V_{\text {glass }}$
$\Rightarrow \frac{20}{3} \gamma_{g l a s s} V_{m e r c u r y}=\gamma_{g l a s s} V$
$\Rightarrow V_{\text {mercury }}=\frac{3 V}{20}$
$\Rightarrow x=3$
Question 7
1 / -0

A steel scale is correct at 0°C, the length of a brass tube measured by it at 40°C is 5m. The correct length of the tube at 0°C is (Coefficients of linear expansion of steel and brass are 11 × 10^-6 /°C and 19 × 10^-6/°C respectively).

A
4.001m

B
5.001m

C
4.999m

D
4.501m

E
4.951m

Solution

Let the length of tube and scale at $0^{\circ} \mathrm{C}$ be I. Then,
Change in measured tube length
$\Delta l=$ Thermal expansion of brass- Thermal expansion of steel
$\Delta l=l \alpha_{B} \Delta T-l \Delta \alpha_{B} \Delta T$
$\Delta l=l \Delta T(\alpha B-\alpha S)$
$\Delta l=l(40-0)\left(19 \times 10^{-6}-11 \times 10^{-6}\right)$
So, new length measured by scale $=l+\Delta l$
$\Rightarrow 5=l+l(40)\left(7 \times 10^{-6}\right)$
$\Rightarrow \frac{5}{1+40 \times 7 \times 10^{-6}}=l$
$\Rightarrow l=4.9986 \approx 4.999 \mathrm{m}$
Option $\mathrm{C}$ is correct.
Question 8
1 / -0

The coefficient of linear expansion 'α' of a rod of length 2m varies with the distance x from the end of the rod as α = α₀ + α₁x where α₀ = 1.76 × 10^-5-1 and α₁ = 1.2 × 10^-6 m^-1-1. The increase in the length of the rod, when heated through 100 is :-

A
2 cm

B
3.76 mm

C
1.2 mm

D
1.5 cm

E
None of these

Solution

$\alpha=\alpha_{0}+\alpha_{1} x$ where $\alpha_{0}=1.76 \times 10^{-5} C^{-1}$
and $\alpha_{1}=1.2 \times 10^{-6} m^{-1} C^{-1}$
and $\Delta T=100^{\circ} C$
$\Delta l=\alpha . d x . \Delta T$
Net extension $l=\int \Delta l=\int \alpha . d x . \Delta T=\int\left(\alpha_{0}+\alpha_{1} x\right) d x . \Delta T$
$l=\Delta T\left(\int_{0}^{L} \alpha_{0} d x+\int_{0}^{L} \alpha_{1} x d x\right)$
$=\Delta T\left(\alpha_{0} L+\frac{\alpha_{1} L^{2}}{2}\right)$
$=100\left(1.76 \times 10^{-5} \times 2+1.2 \times 10^{-6} \times \frac{4}{2}\right)$
$=3.76 \times 10^{-3} \mathrm{m}$
$=3.76 \mathrm{mm}$
Question 9
1 / -0

What is the range of a clinical thermometer?

A
$0-100^{\circ} \mathrm{C}$

B
$32-214^{\circ} F$

C
$0-273^{\circ} \mathrm{C}$

D
$35-42^{\circ} \mathrm{C}$

E
None of these

Solution

A clinical thermometer is used to measure the body temperature and its range is between 35^oC and 42^oC.
Question 10
1 / -0

Thermometer A and B have ice points marked at 15° and 25° and steam points at 75° and J 25° respecti vely. When thermometer A measures the temperature of a bath as 60°, the reading ofB for the same bath is

A
60°

B
70°

C
80°

D
90°

E
100°

Solution

We have
$\frac{A-15}{75-15}=\frac{B-25}{125-25}$ Where $A=$ reading taken by $A$
$\mathrm{B}=$ reading taken by $\mathrm{B}$
$\Rightarrow \frac{60-15}{75-15}=\frac{B-25}{125-25}$
$\Rightarrow \frac{45}{60}=\frac{B-25}{100}$
$\Rightarrow B-25=\frac{3}{4} \times 100$
$\Rightarrow B=100$
$\therefore \quad$ Reading taken by thermometer $\mathrm{B}$ is $100^{\circ}$
Question 11
1 / -0

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
$(1000)^{2}$

E
None of these

Solution

$E=\sigma T^{4} \therefore E \infty(727+273)^{4} \Rightarrow E \infty(1000)^{4}$
Question 12
1 / -0

A steel scale measures the length of a copper rod as $l_{0}$ when both are at $20^{\circ} C$, the calibration temperature for the scale. The scale reading when both are at $40^{\circ} \mathrm{C}$, is

A
$\left(1+20 \alpha_{C}\right) l_{0}$

B
$\left(1+20 \alpha_{S}\right) l_{0}$

C
$\left(\frac{1+20 \alpha_{S}}{1+20 \alpha_{C}}\right) l_{0}$

D
$\left(\frac{1+20 \alpha_{C}}{1+20 \alpha_{S}}\right) l_{0}$

E
None of these

Solution

$l_{o}$ is the reading at $20^{\circ} \mathrm{C}$
Let $l_{1}$ be the scale reading at $20^{\circ} \mathrm{C}$ which gets aligned with the scale at $40^{\circ} \mathrm{C}$
At $40^{\circ} \mathrm{C}$, length of the scale is $l_{o}\left(1+20 \alpha_{c}\right)$
The aligned reading on scale is $l_{1}\left(1+20 \alpha_{s}\right)$
So $l_{o}\left(1+20 \alpha_{c}\right)=l_{1}\left(1+20 \alpha_{s}\right)$
$\Rightarrow l_{1}=\left(\frac{1+20 \alpha_{c}}{1+20 \alpha_{s}}\right) l_{o}$
Question 13
1 / -0

The volume of a metal sphere increases by 0.24% when its temperature is raised by 40. The coefficient of linear expansion of the metal is:

A

$2 \times 10^{-5-1}$

B

$6 \times 10^{-5-1}$

C

$18 \times 10^{-5-1}$

D

$1.2 \times 10^{-5-1}$

E
None of these

Solution

FROM, $v_{T}=v_{0}(1+\gamma \Delta T)$
$\Longrightarrow \quad \frac{V_{T}-V_{0}}{V_{0}}=\gamma \Delta T$
$\frac{0.24}{100}=\gamma 40$
$\gamma=\frac{0.24}{100 \times 40}=6 \times 10^{-5-1}$
Coefficient of linear expansion,
$\alpha=\frac{\gamma}{3}=\frac{6 \times 10^{-5}}{3}=2 \times 10^{-5-1}$
Question 14
1 / -0

If two identical spheres of same radii R out of which one is hollow containing air and the other is solid are heated through the same temperature range, then the change in dimensions of the spheres will be

A
Greater in hollow sphere

B
Greater in Solid sphere

C
Same in both spheres

D
Greater initially in hollow sphere and decreases after certain temperature

E
Greater initially in solid sphere and decreases after certain temperature.

Solution

If two identical spheres of same radii R out of which one is hollow containing air and the other is solid are heated through the same temperature range, then the change in dimensions of the spheres will be greater in Solid sphere.
Question 15
1 / -0

A crystal has a coefficient .Of expansion 13 × 10^-7 in one direction and 231 × 10^-7 in every direction at right angles to it. Then the cubical coefficient of expansion is

A
462 × 10^-7

B
244 × 10^-7

C
475 × 10^-7

D
257 × 10^-7

E
275 × 10^-7

Solution

The cubical expansion is defined as the sum three linear coefficients in three perpendicular directions
$\gamma=\alpha_{1}+\alpha_{2}+\alpha_{3}$
$\Rightarrow 13 \times 10^{-7}+231 \times 10^{-7}+231 \times-7$
$\gamma=475 \times 10^{-7} /^{\circ} \mathrm{C}$
Question 16
1 / -0

The difference between the two temperatures in Celsius scale is 100°. Then the difference between their corresponding Fahrenheit scales is

A
100°F

B
200°F

C
180°F

D
150°F

E
175°F

Solution

In the Celsius scale there are 100 degrees between the freezing point and the boiling point of water compared to 180 degrees in the Fahrenheit scale. This means that 1°C = 1.8°F (check the section about temperature differences below).
Question 17
1 / -0

How many parts does a Celsius scale have?

A
100

B
273

C
180

D
50

E
None of these

Solution

The difference between the lower fixed point and the upper fixed point is 100 parts on the celsius scale.
Question 18
1 / -0

Two bars of same length and same cross-sectional area but of different thermal conductivities $K_{1}$ and $K_{2}$ are joined end to end as shown in the figure. One end of the compound bar is at temperature $T_{1}$ and the opposite end at temperature $t_{2}$ (where $T_{1}>T_{2}$ ). the temperature of the junction is

A
$\frac{K_{1} T_{1}+K_{2} T_{2}}{K_{1}+K_{2}}$

B
$\frac{K_{1} T_{2}+K_{2} T_{1}}{K_{1}+K_{2}}$

C
$\frac{K_{1}\left(T_{1}+T_{2}\right)}{K_{2}}$

D
$\frac{K_{2}\left(T_{1}+T_{2}\right)}{K_{1}}$

E
None of these

Solution

Let L and A be length and area of cross-section of each bar respectively.
∴ Heat current through the bar 1 is
$H_{1}=\frac{K_{2} A\left(T_{1}-T_{0}\right)}{L}$
Here $T_{0}$ is junction temperature.
Heat current through the bar 2 is
$H_{1}=\frac{K_{2} A\left(T_{0}-T_{2}\right)}{L}$
at steady state, $H_{1}=H_{2}$
$\therefore \frac{K_{1} A\left(T_{1}-T_{0}\right)}{L}=\frac{K_{2} A\left(T_{0}-T_{2}\right)}{L}$
$K_{1}\left(T_{1}-T_{0}\right)=K_{2}\left(T_{0}-T_{2}\right)$
$K_{1} T_{1}-K_{1} T_{0}=K_{2} T_{0}-K_{2} T_{2}$
$K_{1} T_{0}-K_{2} T_{0}=K_{1} T_{1}-K_{2} T_{2}$
$T_{0}\left(K_{1}+K_{2}\right)=K_{1} T_{1}+K_{2} T_{2}$
$T_{0}=\frac{K_{1} T_{1}+K_{2} T_{2}}{\left(K_{1}+K_{2}\right)}$
Question 19
1 / -0

Gaps are left between railway tracks because:

A
Gaps give the space to the tracks to expand in summer heat

B
Gaps hold the tracks firmly

C
To produce gentle rhythmic sound when the train moves on the track

D
It is customary to leave the gaps

E
None of these

Solution

The small gaps that are left between the section of the rail that forms each side of the parallel railways for over which railway engines, coaches, trains and trams run. These gaps which are of the order of a few millimeters are provided to allow room for the rails to expand the rise in temperature due to the atmospheric temperature as well as the friction caused by the running of the train. All materials expand when heated and contract when cooled. The extent of expansion and contraction is quite low to be noticed by people or make much difference in most situations. However, since each section of the rail is very long, the expansion due to heating is significant, and unless space is provided for the expansion of rails, the rails will bend to accommodate the increased length. This will make the railway unfit for running trains.
Hence, the correct option is (A).
Question 20
1 / -0

The time period of a pendulum is given by T = $2 π \sqrt{\frac{1}{g}}$ if the temperature rises by △θ and coefficient of linear expansion of the wire is α, find out the change in time period is:

A
α△θxT

B
$\frac{1}{2}$ α△θT

C
2α△θxT

D
$\frac{1}{2}$ α△θ

E
Tα△θ

Solution

Time period of pendulum (simple) $T=2 \pi \sqrt{\frac{l}{g}}$
(l) =initial length of pendulum.
Now, coefficient of linear expansion of a wire $\alpha$ can be defined as,
$\alpha=\frac{\text {increase in length}}{\text {temperature changed} \times \text {initial } \text {length}}$
$\Rightarrow$ Increase in length $=\alpha \times$ (Change in temperature) $\times$ (Initial length)
$\Rightarrow$ Increase in length $=\alpha \times \Delta \theta \times l$
So, new length of wire $l^{\prime}=l+\alpha \triangle \theta l=l(1+\alpha \Delta \theta)$
So, now the time period is $T^{\prime}=2 \pi \sqrt{\frac{l^{\prime}}{g}}$
$\sqrt{l^{\prime}}=\sqrt{l}(1+\alpha \Delta \theta)^{\frac{1}{2}}$
As, $\alpha$ and $\triangle \theta$ are very small, with respect to 1
$\mathrm{So},(1+\alpha \Delta \theta)^{\frac{1}{2}} \approx\left(1+\frac{1}{2} \alpha \Delta \theta\right)$
We neglect the higher order term.
So, $\sqrt{l}=\sqrt{l}\left(1+\frac{1}{2} \alpha \triangle \theta\right)$
and, $T^{\prime}=2 \pi \sqrt{\frac{l}{g}}\left(1+\frac{1}{2} \alpha \Delta \theta\right)=T\left(1+\frac{1}{2} \alpha \Delta \theta\right)$
$\Rightarrow T^{\prime}=T+\frac{1}{2} \alpha \triangle \theta T$
$\Rightarrow T^{\prime}-T=$ Change in time period$=\frac{1}{2} \alpha \triangle \theta T$

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Heat Test - 2

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SHARING IS CARING

Question 1

The room will be warmed up gradually.

The temperature of the room will remain unaffected.

The room will become as cold as the refrigerator.

The room will be cooled slightly.

Solution

Question 2

-40oC

40oC

0oC

32oC

-32oC

Solution

Question 3

γ > 3α

γ < 3α

γ = 3α

γ - 3α

Insufficient information

Solution

Question 4

\(0.85 c c\)

\(0.46 c c\)

\(0.153 c c\)

\(0.05 c c\)

None of these

Solution

Question 5

The difference in the temperatures of the two bodies

The difference in the pressures exerted on the two bodies

The difference in the densities of the two bodies

The difference in the sizes of the two bodies

None of these

Solution

Question 6

3

6

9

12

15

Solution

Question 7

4.001m

5.001m

4.999m

4.501m

4.951m

Solution

Question 8

2 cm

3.76 mm

1.2 mm

1.5 cm

None of these

Solution

Question 9

\(0-100^{\circ} \mathrm{C}\)

\(32-214^{\circ} F\)

\(0-273^{\circ} \mathrm{C}\)

\(35-42^{\circ} \mathrm{C}\)

None of these

Solution

Question 10

60°

70°

80°

90°

100°

Solution

Question 11

\((727)^{4}\)

\((727)^{2}\)

-40^oC

40^oC

0^oC

32^oC

-32^oC

462 × 10^-7

244 × 10^-7

475 × 10^-7

257 × 10^-7

275 × 10^-7

$\frac{1}{2}$ α△θT

$\frac{1}{2}$ α△θ