Tag: measurement and effects of heat

Quizzes (6) Questions (86)

Questions Related to measurement and effects of heat

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

A given amount of gas occupies 1000cc at 27$^{0}$ C and 1200cc and 87$^{0}$ c. What is its volume  coefficient of expansion

  1. $\frac{1}{273}^{0}C^{-1}$

  2. $\frac{1}{173}^{0}C^{-1}$

  3. $173^{0}C^{-1}$

  4. $273^{0}C^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know , $\alpha =\frac { { V } _{ 2 }-{ V } _{ 1 } }{ { V } _{ 1 }{ t } _{ 2 }-{ V } _{ 2 }{ t } _{ 1 } } $
Substituting the values ${ V } _{ 2 }=1200cc$ , ${ V } _{ 1 }=1000cc$, ${ t } _{ 2 }={ 87 }^{ \circ  }C$, ${ t } _{1}={ 27 }^{ \circ  }C$.
$\therefore \alpha =\frac { 200 }{ \left( 87000-32400 \right)  } $
$\therefore \alpha ={ \frac { 1 }{ 273 }  }^{ \circ  }{ C }^{ -1 }$

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

The coefficient of volume expansion of liquid is $\gamma$. The fractional change in its density for $\triangle T$ rise in temperature is ?

  1. $\gamma \triangle T$

  2. $\dfrac{\triangle T}{\gamma}$

  3. $1+\gamma \triangle T$

  4. $1-\gamma \triangle T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

On thermal expansion,

Volumetric expansion is given by
$V=V _0(1+\gamma \Delta T)$. . . . . . . .(1)
We know that, density, $d=\dfrac{mass}{volume}$
$d=\dfrac{m}{V}$
where, $m=$ constant
$d\propto \dfrac{1}{V}$
Density of the liquid varies as

$d=d _0(1+\gamma \Delta T)$
$d=d _0+d _0\gamma \Delta T$
Fractional change in density is 
$\dfrac{d-d _0}{d _0}=\gamma \Delta T$
$\dfrac{\Delta d}{d _0}=\gamma \Delta T$
The correct option is A.

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

$1$ mole of a gas with $\gamma =\dfrac{7}{5}$ is mixed with $1$ mole of gas with $\gamma =\dfrac{5}{3}$, the value of $\gamma$ of the resulting mixture of.

  1. $\dfrac{7}{5}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{3}{2}$

  4. $\dfrac{12}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ Y } _{ mis }=\cfrac { { n } _{ 1 }C{ \rho  } _{ 1 }+{ n } _{ 2 }C{ \rho  } _{ 2 } }{ { n } _{ 1 }C{ \gamma  } _{ 1 }+{ n } _{ 2 }C{ \gamma  } _{ 2 } } $

${ C\rho  } _{ 1 }=\cfrac { 5 }{ 2 } R$ then its $C{ v } _{ 1 }=\cfrac { 3 }{ 2 } R$
Because ${ C } _{ \rho  }-{ C } _{ v }=R$
for diatomic gas ${ C\rho  } _{ 2 }=\cfrac { 7R }{ 2 } $ then ${ Cv } _{ 2 }=\cfrac { 5 }{ 2 } R$
${ Y } _{ mis }=\cfrac { { n } _{ 1 }\times \cfrac { 5 }{ 2 } R+{ n } _{ 2 }\times \cfrac { 7 }{ 2 } R }{ { n } _{ 1 }\times \cfrac { 3 }{ 2 } R+{ n } _{ 2 }\times \cfrac { 5 }{ 2 } R } $
Here ${ n } _{ 1 }={ n } _{ 2 }=1$
${ Y } _{ mis }=\cfrac { 3 }{ 2 } $

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

If $T$ represent the absolute temperature of an ideal gas, the volume coefficient of thermal expansion at constant pressure, is :

  1. $T$

  2. $T^2$

  3. $1/T$

  4. $1/T^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From the definition of $\gamma _p$ 


We have $V _t=V _0(1+\gamma _pt)$..........(1) 

Again from Charle's law, $V _t=V _0(1+\dfrac{1}{T}t)$...........(2)  

Comparing (1) and (2), 

$\gamma _p=\dfrac{1}{T}$

Hence,option C is correct.

Multiple choice physics measurement and effects of heat sources of heat introduction to heat measuring temperature

A man wishes to fit an aluminium ring on steel rod of 1 cm diameter and found it is 0.01 cm smaller in diameter. How much should the on temperature be raised before it just slips on the
$(\alpha _A = 25 \times 10^{-6}/^oC;$ $10 \times 10^{-6}/^oC)$

  1. $40^oC$

  2. $50^oC$

  3. $25^oC$

  4. $60^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${l _2} = {l _1}\left( {1 + {\alpha _A}\Delta T} \right)$

${l _2} = {l _1} + {l _1}{\alpha _A}\Delta T$

$\displaystyle {{{l _2} - {l _1}} \over {{l _1}}} = {\alpha _A}\Delta T$

$\displaystyle {{0.01 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 2}} \times 25 \times {{10}^{ - 6}}}} = \Delta T$

$\Delta T = {40^0}C$