Tag: introduction to temperature

Questions Related to introduction to temperature

The resistance of a platinum wire of a platinum resistance thermometer at the ice point is $5 \Omega$ and at steam point is $5.4 \Omega$. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $6.2 \Omega$. Find the temperature of the hot bath.

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. $300^\circ C$

  2. $30^\circ C$

  3. $3000^\circ C$

  4. $300 \ K$

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Correct Option: A
Explanation:
Given,

Resistance of platinum wire at ice point, $R _0=5\,\Omega$

Resistance of platinum wire at hot bath $R _H=6.2\,\Omega$

Temperature at hot bath $T _H=?$

We have,

$R _T=R _0[1+\alpha(T-T _0)]$

$\implies R _{100}=R _0[1+\alpha(T _{100}-T _0)]$

$\implies 5.4=5[1+\alpha(100-0)]$

$\implies \dfrac{5.4}{5}-1=100\alpha$

$\implies \alpha=\dfrac{1}{1250}  \, ^0 C^{-1}$

Also,

$R _H=R _0[1+\alpha (T _H-T _0)]$

That is,

$6.2=5[1+\dfrac{1}{1250}(T _H-0)]$

$\dfrac{6.2}{5}-1=\dfrac{1}{1250}\times T _H$

$\implies T _H=300^0 C$


A thermometer uses 

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. water

  2. mercury

  3. air

  4. none

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Correct Option: B

45 gm of alcohol are needed to completely fill up a weight thermometer at $15^{\circ}C$. Find the weight of alcohol which will overflow when the weight thermometer is heated to $33^{circ}C$.
(Given ${ \gamma  } _{ a }=121\times { 10 }^{ -5 }{ { \circ  } _{ C } }^{ -1 }$

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. 0.96 gm

  2. 0.9 gm

  3. 1 gm

  4. 2 gm

Show answer
Correct Option: C

Consider two thermometers $T _1$ and $T _2$ of equal length which can be used to measure temperature over the range $\theta _1$ and $\theta _2$. $T _1$ contains mercury as thermometric liquid while $T _2$ contains bromine. The volumes of the two liquids are the same at the temperature $\theta _1$. The volumetric coefficients of expansion of mercury and bromine are $18\times 10^{-5}K^{-1}$ and $108\times 10^{-5}K^{-1}$, respectively. The increase in length of each liquid is the same for the same increase in temperature. If the diameters of the capillary tubes if the two thermometers are $d _1$ and $d _2$ respectively, then the ratio $d _1:d _2$ would be closest to.

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. $6.0$

  2. $2.5$

  3. $0.5$

  4. $0.4$

Show answer
Correct Option: D
Explanation:

Increase in length of each liquid is same 

$\dfrac{\Delta V _{hg}}{\pi d _1^2}=\dfrac{\Delta V _{br}}{\pi d _2^2}$
$\dfrac{\Delta V _{hg}\Delta\theta}{\pi d _1^2}=\dfrac{\Delta V _{br}\Delta\theta}{\pi d _2^2}$
$\dfrac{d _12}{d _2^2}=\dfrac{\gamma _{hg}}{\gamma _{br}}=\dfrac{1}{6}$
$\dfrac{d _1}{d _2}=0.4$

$\begin{array} { l } { \text { Energy required to dissociate } 4 \mathrm { g } \text { of gaseous } } \ { \text { hydrogen into free gaseous atoms is } 208 \mathrm { Kcal {at}  }  } \ {  25 ^ { \circ } \mathrm { C } \text { . The bond energy of } \mathrm { H } - \mathrm { H } \text { bond will be : } } \end{array}$ .

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. $1.04Kcal$

  2. $10.4Kcal$

  3. $104Kcal$

  4. $1040Kcal$

Show answer
Correct Option: C
Explanation:
Given heat of atmosphere $40=260\ Kcal$
$2H _{2} \rightarrow 4H$
$\triangle H = 208\ Kcal$
$20 \rightarrow 1\ mole$
$40 \rightarrow 2 \ mole$
$Hene \ 2 H-H$ bonds area brown $bg $
$20\ kcal $ energy so in order to break $1\ H-H$ bound we required $\dfrac{208}{2}= 104\ Kcal$
Hence the bond energy of $H-H$ bound will be $=104\ kcal$

Which of the following statements is correct?

physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

  1. Air escaping from a punctured tyre feels cold

  2. When a gas under high pressure is permitted to expand into a region of low pressure, it gains in temperature

  3. The reading on a thermometer immersed in boiling water varies as the heat increases or decreases above the boiling point

  4. None of the above statements is correct

Show answer
Correct Option: A
Explanation:

Option (A) is correct. 

Reason - The air is contained at high pressure in the tube. When it escapes through a small hole, it suddenly expands. A large amount of heat is absorbed in the process of expansion resulting in considerable fall in its temperature. This is why the escaping air feels cold.

150 g of ice is mixed with 100 g of water at temperature $80^oC$. The latent heat of ice is 80 ca/g and the specific heat of water is $1 cal/g-^oC$. Assuming no heat loss to the environment, the amount of ice which does not melt is

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. 100 g

  2. 0 g

  3. 150 g

  4. 50 g

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Correct Option: D
Explanation:

Heat given by water $=100\times 1\times 800=8000 cal$
Heat taken by ice $=8000 cal=m\times 80$
$m=100 gm$
So amount of ice which does not melt $=150-100=50 gm$.

The following three objects (1) a metal tray, (2) a block of wood,and (3) a wooden cap are left in a closed room overnight. Next day the temperature of each is recorded as $T _1, T _2$ and $T _3$ respectively. The likely situation is 

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. $ T _1 = T _2 = T _3 $

  2. $ T _3 > T _2 > T _1 $

  3. $ T _3 = T _2 > T _1 $

  4. $ T _3 > T _2 = T _1 $

Show answer
Correct Option: A
Explanation:

All three will be inthermal equilibrium with air of room. so temprature of the three will be same

A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature t$ _x$ in this scale is related to the absolute temperature T by t$ _x$ = 3T + 300. If the specific heat of a material using this unit is 1400 J kg$^{-1}$ X${^-1}$ its specificne in the S.I. system of units is

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. 4200 J kg$^{-1}$ K$^{-1}$

  2. 1400 J kg$^{-1}$ K$^{-1}$

  3. 466.7 J kg$^{-1}$ K$^{-1}$

  4. impossible to determine from the information provided

Show answer
Correct Option: A
Explanation:

$t _x = 3 T + 300                 Q = ms \Delta 0$
$\Delta t _x = 3 \Delta T$
$\displaystyle S = \frac{Q}{m \Delta \theta}                 S = \frac{Q}{m \Delta \theta}$
Since, unit of 8 is joule in both system
            X                                    T
$m = m _0 kg$                   $m _0 kg$
$Q = Q _0 J$                       $Q _0  J$
$\Delta t _x$                               $\Delta T$
$S _x = \displaystyle \frac{Q _0}{m _0 \Delta t _x} = 1400     S _r = \frac{Q _0}{m _0 \Delta T} = \frac{3 Q}{m _0 \Delta t _x}$
$S _r = 3 \times 1400 = 4200 J-kg^{-1} K^{-1}$

Two glass tumblers have been stuck together (one into the other). They can be seperated by

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. Placing hot water in the inner tumbler

  2. Placing the tumblers in cold water

  3. Placing the outer tumbler in hot water

  4. Hamming them vigorously

Show answer
Correct Option: C
Explanation:

The correct answer is option(C).

They can be separated by placing cold water in inside tumbler and placing the outer tumbler in hot water. This way it creates a temperature difference and the glass tumblers get separated easily.