Tag: heat - measurement

Questions Related to heat - measurement

A piece of lead falls from height of $100m$ on a fixed non-conducting slab which brings it to rest. if the specific heat of lead is $30.6 cal/kg ^ \circ C $ the increase in temperature of the slab immediately after collision is  

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

  1. ${6.72^ \circ }C$

  2. ${7.62^ \circ }C$

  3. ${5.62^ \circ }C$

  4. ${8.72^ \circ }C$

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Correct Option: B

A physicist says "a body contains $10\ joule$ heat" but a physics learner says "this statement is correct only when the body is in liquid state". Mark correct option or options :

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

  1. physicist statement is correct

  2. physics learner's statement is correct

  3. both statements are correct

  4. both statements are wrong

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Correct Option: D

Two identical rods of a mental are welded in series then 20 cal of heat flows through them in  4 minute. If the rods are welded in parallel then same amount of heat will flow in 

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

  1. 1 minute

  2. 2 minute

  3. 4 minute

  4. 15 minute

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Correct Option: A
Explanation:

$dq/dt = [(AK)/2x][T2-T1] = 20/4 = 5.$

 $Therefore ( AK/x)(T2-T1) = 10$

When they are one above the other,

$dq/dt = 2Ak/x][T2-T1] =20/t$

Therefore$t = 2\times10  = 20/t$

Therefore$ t = 1 min$

A liter of air at $20^oC$ is heated until both the pressure and the volume are tripled, what is the tempertare then.

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

  1. $2637^oK$

  2. $927^oK$

  3. $200^oK$

  4. $977^oK$

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Correct Option: A
Explanation:

Applying the formula

$PV=nRT$
$\dfrac { { P } _{ 1 } }{ { V } _{ 1 } } =\dfrac { { P } _{ 2 } }{ { V } _{ 2 } } =\dfrac { { T } _{ 1 } }{ { T } _{ 2 } } $    [$R$ is constant]
Let ${ P } _{ 1 }=P$  and ${ V } _{ 1 }=V$
As given ${ P } _{ 2 }=3P$   ${ V } _{ 2 }=3V$
${ T } _{ 1 }={ 20 }^{ 0 }C=20+2+3=293$
${ T } _{ 2 }=?$
$\dfrac { PV }{ 3P\times 3V } =\dfrac { 293 }{ { T } _{ 2 } } $
$\dfrac { 1 }{ 9 } =\dfrac { 293 }{ { T } _{ 2 } } $
${ T } _{ 2 }=293\times 9=2637$
$\therefore$    New temperature $=2637$.