Question

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

• A. $4.2\times {10}^3$
• B. $3.0\times {10}^3$
• C. $1.2\times {10}^3$
• D. $1.5\times {10}^3$

Answer 1

The correct option is D $1.5\times {10}^3$

Specific heat capacity of water, $S=4200$ $\frac{J}{kg\times K}$

Given : For new temperature scale, Ice point $=25X$

Steam point $=305X$

Difference in the steam point and ice point is equal to $100K$

$\therefore$ $305X-25X=100K$

$⟹1K=2.8X$

This means $1$ Kelvin is equal to $2.8$ $X$

$\therefore$ $S=4200$ $\frac{J}{kg\times K}=$ $4200$ $\frac{J}{kg\times 2.8X}=1500$$\frac{J}{kg\times X}$

$⟹$ Specific heat capacity of water is $1.5\times {10}^3$ $J$ $k{g}^{-1}$ $X^{-1}$