Tag: introduction to temperature

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Questions Related to introduction to temperature

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

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 :

  1. physicist statement is correct

  2. physics learner's statement is correct

  3. both statements are correct

  4. both statements are wrong
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat is a form of energy in transit between systems due to a temperature difference; it is not a state function or a property contained within a body. Therefore, saying a body 'contains' heat is physically incorrect regardless of its state.

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

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 

  1. 1 minute

  2. 2 minute

  3. 4 minute

  4. 15 minute
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

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

  1. $2637^oK$

  2. $927^oK$

  3. $200^oK$

  4. $977^oK$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=2637$.