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 kg${}^{-1}$ X${}^{-}1$ its specificne in the S.I. system of units is

• A. 4200 J kg${}^{-1}$ K${}^{-1}$
• B. 1400 J kg${}^{-1}$ K${}^{-1}$
• C. 466.7 J kg${}^{-1}$ K${}^{-1}$
• D. impossible to determine from the information provided

Answer 1

The correct option is A 4200 J kg${}^{-1}$ K${}^{-1}$

$t_x=3T+300Q=ms\Delta 0$
$\Delta t_x=3\Delta T$
$S=\frac{Q}{m\Delta \theta }S=\frac{Q}{m\Delta \theta }$
Since, unit of 8 is joule in both system
X T
$m=m_{0}kg$ $m_{0}kg$
$Q=Q_{0}J$ $Q_{0}J$
$\Delta t_x$ $\Delta T$
$S_x=\frac{Q_0}{m_0\Delta t_x}=1400S_r=\frac{Q_0}{m_0\Delta T}=\frac{3Q}{m_0\Delta t_x}$
$S_r=3\times 1400=4200J-k{g}^{-1}{K}^{-1}$