Thermal Properties Of Matter(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

THERMAL PROPERTIES OF MATTER MCQs

Total Questions : 32 | Page 1 of 4 pages

It was a bright and breezy day in New York, the air was a cool 20 degrees Celsius, when Tony Stark, a.k.a the Iron Man's day took a dramatic turn as he got news of Mandarin's attack in the windy city of Chicago, and decided to immediately fly over there, in his iron body-armor, whose cavity had enough space to fit a man of 4,984 $c m^{3}$ in volume, but not more.

Being a smart man, he did a quick check of two important things - his own body volume, which he found was currently 4.980 $c m^{3}$ , and the temperature in Chicago, which was a cold 8 degrees Celsius that day. Knowing that the coefficient of volume expansion for iron, $γ$ , is $33.3 \times 10^{- 6}$ $/^{\circ} C$ , he decided to go. Check if that was a smart decision, by finding out whether the new volume of the iron suit when he reaches Chicago will crush him or not.

$4.982 c m^{3}$ (he is going to be alright)
$4.979 c m^{3}$ (it's going to be an uncomfortable encounter)
$4.938 c m^{3}$ (he will have to take his suit off and call the Hulk)
$4.980 c m^{3}$ (perfect fit!).

Discuss Question

Answer: Option A. ->

4.982 c m^{3}

(he is going to be alright)
:
A

Mr. Stark's suit is made of iron, which should contract upon cooling with a coefficient $γ$ = $33.3 \times 10^{- 6}$ $/^{\circ} C$ . As he flies from New York to Chicago, there occurs a temperature drop of $12^{\circ} C$ .

$∴$ $Δ T$ = $T_{C h i c a g o}$ - $T_{N Y}$ = ( $8^{\circ} C$ - $20^{\circ} C$ ) = $- 12^{\circ} C$

The volume, $V_{N Y}$ , of the suit's cavity in New York = $4, 984 c m^{3}$ .

$∴$ The volume of the cavity in Chicago will be -

$V_{C h i c a g o}$ = $V_{N Y} (1 + γ Δ T)$

= $4.984 [1 + 33.3 \times 10^{- 6} \times (- 12)] c m^{3}$

$\approx$ $4.982 c m^{3}$ .

Awesome! The suit will still fit Tony Stark comfortably in Chicago, so he can just concentrate on Mandarin's onslaught. I feel much less worried about Chicago now!

Question 2.

Browsing through Mr. Fox's old scientific notes, Bruce Wayne encounters a secret temperature scale, which Mr. Fox called Z, in order to keep the Batmobile's technology a secret. On that scale, the freezing and the boiling points of water were recorded to be -7⁰Z and 78⁰Z, respectively. The records instruct to maintain the coolant temperature at -24⁰Z for top speed. Mr. Wayne, naturally comfortable with the Fahrenheit scale due to his American upbringing, needs to calculate the coolant temperature in ⁰F. Help him choose the correct value.

10℉
-4℉
5℉
0

Discuss Question

Answer: Option B. -> -4℉
:
B

Let's list what we know -

For any general scale with lower and upper fixed points labeled as T_L and T_u, and a measured temperature T, the ratio will be the same, no matter what scale it is.

Extending that to ⁰Fand ⁰Z(in question), we have -

Question 3.

A platinum sphere floats in mercury. Find the percentage change in the fraction of volume of sphere immersed in mercury when the temperature is raised by $80^{\circ} C$ (Volume expansivity of mercury is $182 \times 10^{- 6} /^{\circ} C$ and linear expansivity of platinum is $9 \times 10^{- 6} /^{\circ} C$ )

$1.24$
$1.38$
$2.48$
$2.76$

Discuss Question

Answer: Option A. ->

1.24

:
A

$\frac{V^{'}}{V}$ = $\frac{ρ}{σ}$ = Fraction of volume of sphere submerged = $η$ (say)

To find $%$ change in $η$ , i.e., $(\frac{η^{'}}{η} - η) \times 100$

Or $(\frac{η^{'}}{η} - η) \times 100$ = $[\frac{(\frac{ρ}{σ})^{'}}{(\frac{ρ}{σ})} - 1] \times 100$

= $[\frac{(1 + γ_{m} Δ T)}{(1 + γ_{p} Δ T)}] \times 100 ≃ (γ_{m} - γ_{p}) Δ T \times 100$

= $(182 - 27) \times 10^{- 6} \times 100$ = $1.24 %$

Question 4.

A system X is neither in thermal equilibrium with Y, nor with Z. The systems Y and Z -

Must be in thermal equilibrium
Cannot be in thermal equilibrium
May be in thermal equilibrium
None of these

Discuss Question

Answer: Option C. -> May be in thermal equilibrium
:
C

Let the temperatures of X, Y and Z be T_x, T_y and T_z, respectively.

Now,

T_x ≠ T_y,

and

T_x ≠ T_z.

This does not tell us anything about T_y and T_z, except that they are not equal to T_x.

So, T_y = T_z, and T_y ≠ T_z, both are possibilities!

Hence, option (c) is correct.

Question 5.

Two rods of lengths $l_{1}$ and $l_{2}$ are made of materials whose coefficients of linear expansion are $α_{1}$ and $α_{2}$ . If the difference between the two lengths is independent of temperature, then

$\frac{l_{1}}{l_{2}} = \frac{α_{1}}{α_{2}}$
$\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}$
$l_{2}^{2} α_{1} = l_{1}^{2} α_{2}$
$\frac{α_{1}^{2}}{l_{1}} = \frac{α_{2}^{2}}{l_{2}}$

Discuss Question

Answer: Option B. ->

\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}

:
B

L_{1} = l 1 [1 + α_{1} Δ T] L_{2} = l 2 [1 + α_{2} Δ T] L_{1} + L_{2} = (l_{1} + l 2) + (Δ T) (l_{1} α_{1} - l_{2} α_{2}) I f L_{1} - L_{2} i s t o b e i n d e p e n d e n t o f Δ T, t h e n l_{1} α_{1} - l_{2} α_{2} = 0 \Rightarrow \frac{l_{1}}{l 2} = \frac{α_{1}}{α_{2}} \frac{l_{1}}{l 2} = \frac{α_{2}}{α_{1}}

Question 6.

You can use a mercury thermometer to accurately infer whether mercury expands linearly with temperature

True
False
5℉
0

Discuss Question

Answer: Option B. -> False
:
B

No! You cannot use the thermometer to verify the linear expansion of mercury. This is because the temperature scale itself is defined by making equal markings between 0⁰ C and 100⁰ C, assuming that mercury expands linearly.

In fact, scientists have now established using statistical physics and other concepts that mercury does not expand exactly in a linear fashion.

Before all this, the only way to guess if a thermometer fluid is a good choice (i.e, expands linearly) was to observe which of them gave consistent results.

Question 7.

In a closed container of constant volume, the pressure of water exhibits an interesting dependence on temperature -

The point D (273.16 K,0.006 atm) is called the triple point, Tp, of water, where all there phases coexist. What is the observed phase change when the temperature is increased from -10⁰C to +10⁰C, while maintaining a constant pressure of 0.006 atm?

Solid → Liquid
Liquid → Gas
Solid → Gas
Gas → Liquid

Discuss Question

Answer: Option C. -> Solid → Gas
:
C

Along the constant pressure line of P=0.006, we have ice (solid H₂O) on the left of the triple point, and steam (gaseous H₂O) on the right.

Thus, as we increase the temperature from below Tp to above Tp, we observe a solid → gas phase transition, i.e., we see ice directly turning into gas without going through the water phase. Magical!

Question 8.

Steam at $100^{\circ} C$ is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at $15^{\circ} C$ till the temperature of the calorimeter and its contents rise at $80^{\circ} C$ . The mass of the stem condensed in kilogram is

0.130
0.065
0.260
0.135

Discuss Question

Answer: Option A. -> 0.130
:
A
Heat lost by steam = Heat gained by water + calorimeter

∴ m L + m s (100 - 80) = 1.12 \times s \times (80 - 15)

m [540 + (1 \times 20)] = 1.12 \times 1 \times 65

m = \frac{1.12 \times 1 \times 65}{560} = \frac{65}{500} k g o r m = 0.13 k g

Question 9.

A lead bullet strikes against a steel armor plate with the velocity of 300 meter/second. If the bullet, assuming that the impact is perfectly inelastic find the rise in temperature of the bullet. Assume that the heat produced is shared equally between the bullet and the target, specific heat of lead = $0.03 c a l / g /^{\circ} C$

${178.6}^{\circ} C$
${78.6}^{\circ} C$
${2780.6}^{\circ} C$
${100.6}^{\circ} C$

Discuss Question

Answer: Option A. ->

{178.6}^{\circ} C

:
A
Suppose m be the mass of the bullet and T be the rise in temperature.
Heat produced in the bullet

H = m \times s \times T = m \times 0.03 \times T c a l

Kinetic energy of the bullet

= \frac{1}{2} m v^{2} = \frac{1}{2} m (300 \times 10^{2})^{2} e r g s

Half of this energy goes into the bullet on impact, therefore

W = \frac{1}{2} \times [\frac{1}{2} m (300 \times 10^{2})^{2}] = \frac{1}{4} m \times 9 \times 10^{8} e r g s U s i n g t h e f o r m u l a W = J H, w e h a v e \frac{1}{4} m \times 9 \times 10^{8} = (4.2 \times 10^{7}) (m \times 0.03 \times T) T = \frac{9 \times 10^{8}}{4 \times (4.2 \times 10^{7}) \times 0.03} = {178.6}^{\circ} C

Question 10.

A lump of $0.10 k g$ of ice at $- 10^{\circ} C$ is put in $0.15 k g$ of water at $20^{\circ} C$ . How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water $= 1 k c a l k g^{- 1}$ , specific heat of ice $= 0.50 k c a l k g^{- 1}$ while its latent heat $= 80 k c a l k g^{- 1}$ )