Question

A scientist proposes a new temperature scale in which the ice point is $25\, X$ ($X$ is the new unit of temperature) and the steam point is $305\, X$. The specific heat capacity of water in this new scale is (in $J\,kg^{-1}\, X^{-1}$)

• A. $4.2×10^3$
• B. $3.0×10^3$
• C. $1.2×10^3$
• D. $1.5×10^3$

Answer 1

Answer: (D)

Given, $305 X -25\, X =100^{\circ} C$
$(\because X$ is the new unit of temperature)
$(305-25) X =100^{\circ} C $
$\Rightarrow 280 X =100^{\circ} C $
$\therefore 1^{\circ} C =2.8 \,X$
The specific heat capacity of water
$ =4200 \frac{\text { joule }}{ kg ^{\circ} C } =4200 \times \frac{\text { joule }}{ kg \times 2.8 \,X } $
$=1500 \,J / kg - X $
$ =1.5 \times 10^{3}\, J \,kg ^{-1}\, X ^{-1}$