A scientist proposes a new temperature scale | vedclass

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(D) The ice point on the Celsius scale is $0^{\circ} C$ and the steam point is $100^{\circ} C$. In the new scale, the ice point is $25 X$ and the stea

Vedclass · Accepted Answer

$1.5 	imes 10^{3}$

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(D) The ice point on the Celsius scale is $0^{\circ} C$ and the steam point is $100^{\circ} C$. In the new scale, the ice point is $25 X$ and the steam point is $305 X$. The temperature interval of $100^{\circ} C$ corresponds to $(305 - 25) X = 280 X$. Therefore, a change of $1^{\circ} C$ is equivalent to a change of $\frac{280}{100} X = 2.8 X$. This implies $1^{\circ} C = 2.8 X$, or $1 X = \frac{1}{2.8}^{\circ} C$. The specific heat capacity of water is $c = 4200 \ J kg^{-1} (^{\circ} C)^{-1}$. Substituting the relation $1^{\circ} C = 2.8 X$ into the unit: $c = 4200 \ J kg^{-1} (2.8 X)^{-1} = \frac{4200}{2.8} J kg^{-1} X^{-1} = 1500 \ J kg^{-1} X^{-1}$. Thus, $c = 1.5 	imes 10^{3} J kg^{-1} X^{-1}$.

$A$ 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 $J kg^{-1} X^{-1}$):

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