Thermal Properties Of Matter(11th And 12th > Physics ) Questions and Answers
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 cm3 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 cm3, 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 × 10−6/∘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.
Explanation:-
Answer: Option A. -> 4.982 cm3 (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 × 10−6/∘C. As he flies from New York to Chicago, there occurs a temperature drop of 12∘C.
∴ΔT = TChicago - TNY = (8∘C - 20∘C) = −12∘C
The volume, VNY, of the suit's cavity in New York = 4,984 cm3.
∴ The volume of the cavity in Chicago will be -
VChicago = VNY(1+γΔT)
= 4.984[1+33.3 × 10−6 × (−12)]cm3
≈ 4.982 cm3.
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!
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 -70ZÂ and 780Z, respectively. The records instruct to maintain the coolant temperature at -240ZÂ 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.
Explanation:-
Answer: Option A. -> 1.24:
A
V′V = ρσ = Fraction of volume of sphere submerged = η(say)
To find % change in η, i.e., (η′η−η)× 100
Or (η′η−η)× 100 = [(ρσ)′(ρσ)−1]×100
= [(1+γmΔT)(1+γpΔT)]×100≃(γm−γp)ΔT×100
= (182−27)×10−6×100 = 1.24%
Explanation:-
Answer: Option C. -> May be in thermal equilibrium:
C
Let the temperatures of X, Y and Z be Tx, Ty and Tz, respectively.
Now,
Tx ≠ Ty,
and
Tx ≠ Tz.
This does not tell us anything about Ty and Tz, except that they are not equal to Tx.
So, Ty = Tz, and Ty ≠ Tz, both are possibilities!
Hence, option (c) is correct.
Explanation:-
Answer: Option B. -> l1l2=α2α1:
B
L1=l1[1+α1ΔT]L2=l2[1+α2ΔT]L1+L2=(l1+l2)+(ΔT)(l1α1−l2α2)If L1−L2 is to be independent of ΔT,thenl1α1−l2α2=0⇒l1l2=α1α2 l1l2=α2α1
Explanation:-
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.
In a closed container of constant volume, the pressure of water exhibits an interesting dependence on temperature -Â
The point DÂ (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 -100CÂ to +100C, while maintaining a constant pressure of 0.006 atm?Â
Explanation:-
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!
Explanation:-
Answer: Option A. -> 0.130:
A
Heat lost by steam = Heat gained by water + calorimeter
∴ mL+ms(100−80)=1.12×s×(80−15)
or m[540+(1×20)]=1.12×1×65
or m=1.12×1×65560=65500 kg or m=0.13 kg
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 cal/g/∘C
Explanation:-
Answer: Option A. -> 178.6∘C:
A
Suppose m be the mass of the bullet and T be the rise in temperature.
Heat produced in the bullet
H=m×s×T=m×0.03×T cal
Kinetic energy of the bullet
=12mv2=12m(300×102)2ergs
Half of this energy goes into the bullet on impact, therefore
W=12×[12m(300×102)2]=14m×9×108 ergsUsing the formula W=JH,we have14m×9×108=(4.2×107)(m×0.03×T)T=9×1084×(4.2×107)×0.03=178.6∘C
Explanation:-
Answer: Option A. -> Ice is 68.75gm, and water is 181.25gm at 0∘C.:
A
The heat, which 0.15kg of water can release when its temperature is changed from 20∘C to 0∘C.
Q1=mWsWΔTW
where, mW is the mass of the water, sW is the specific heat of water and ΔTW is the change in temperature of water.
Given, mW=0.15 kg, sW=1 kcal kg−1.
Q1=0.15×(1×103)×(20−0)
=3000cal ...(1)
Now, heat absorbed by 0.10kg of ice at −10∘C to increase its temperature to 0∘C.
Q2=miSiΔTi
where mi is the mass of the ice, si is the specific heat of ice and ΔTi is the change in temperature of ice.
Given, mi=0.10 kg, si=0.50 kcal kg−1
Q2=0.10×(0.5×103)×[0−(−10)]
=500cal
So, remaining heat,
Q=Q1−Q2=3000−500=2500 cal
Now as latent heat of ice is l=80kcalkg−1, the remaining heat will melt ice only
L=Qm⇒m=Qm=250080=31.25 g of ice.
Initial amount of ice =0.10 kg=100 g
So, the remaining ice,
=100−31.25=68.75g
Initial amount of water =0.15 kg=150 g
Therefore, total water =150+31.25=181.25g
The temperature of the system will be 0∘C.