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Standard VI

Physics

Celsius Scale

A new tempera...

Question

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 $k g^{- 1} K^{- 1}$ its specific heat in the S.1. system of units is :

4200 J k g^{- 1} K^{- 1}

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1400 J k g^{- 1} K^{- 1}

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466.7 J k g^{- 1} K^{- 1}

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Impossible to determine from the information provided

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Solution

The correct option is A $4200 J k g^{- 1} K^{- 1}$
$t_{x} = 3 T + 300$
$δ (t_{x}) = 3$
Thus, for every one degree change in the SI system, there is a 3 degree change in the new system. Hence, the specific heat in SI system= 1400*3= 4200J/kgK

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Similar questions

Q. 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}

^{-} 1

its specificne in the S.I. system of units is

50

g of copper is heated to increase its temperature by

10^{o}

C. If the same quantity of heat is given to

10

gm of water, the rise in temperature is (specific heat of copper

= 420

k g^{- 1} K^{- 1}

, specific heat of water

= 4200

k g^{- 1} K^{- 1}

The specific heat of a metal at low temperatures varies according to $S = a T^{3}$ where $a$ is a constant and $T$ is the absolute temperature. The heat energy needed to raise unit mass of the metal from $T = 1 K$ to $T = 2 K$ is

Q. If the temperature of 10 kg of water is raised by

10^{o} C

, then the heat energy supplied to water is _________ . (specific heat capacity of water is

4200 J k g^{- 1} K^{- 1}

)

50 g

of copper is heated to increase its temperature by

10^{\circ} C

. If the same quantity of heat is given to

10 g

of water, then the rise in its temperature (in

^{\circ} C

) is:
(Specific heat of copper

= 420 J k g^{- 1} K^{- 1}

,
Specific heat of water

= 4186 J k g^{- 1} K^{- 1}

)

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The correct option is A 4200Jkg−1K−1tx=3T+300δ(tx)=3Thus, for every one degree change in the SI system, there is a 3 degree change in the new system. Hence, the specific heat in SI system= 1400*3= 4200J/kgK

The correct option is A $4200 J k g^{- 1} K^{- 1}$
$t_{x} = 3 T + 300$
$δ (t_{x}) = 3$
Thus, for every one degree change in the SI system, there is a 3 degree change in the new system. Hence, the specific heat in SI system= 1400*3= 4200J/kgK