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 :
$4200 J k g^{- 1} K^{- 1}$
$1400 J k g^{- 1} K^{- 1}$
$466.7 J k g^{- 1} K^{- 1}$
Impossible to determine from the information provided

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 :
$4200 J k g^{- 1} K^{- 1}$
$1400 J k g^{- 1} K^{- 1}$
$466.7 J k g^{- 1} K^{- 1}$
Impossible to determine from the information provided

Toppr Admin · Accepted Answer

Tx-3T-300-x3B4-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

tx=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

$t_{x} = 3 T + 300$
$δ (t_{x}) = 3$
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