The temperature of a body on the Kelvin scale | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between the Fahrenheit scale $(F)$ and the Kelvin scale $(K)$ is given by the formula: $\frac{F - 32}{9} = \frac{K - 273.15}{5}$.

Vedclass · Accepted Answer

$574.25$

Vedclass · Answer

(C) The relationship between the Fahrenheit scale $(F)$ and the Kelvin scale $(K)$ is given by the formula: $\frac{F - 32}{9} = \frac{K - 273.15}{5}$.Given that the numerical value is the same on both scales,we set $F = x$ and $K = x$.Substituting these into the equation: $\frac{x - 32}{9} = \frac{x - 273.15}{5}$.Cross-multiplying gives: $5(x - 32) = 9(x - 273.15)$.$5x - 160 = 9x - 2458.35$.Rearranging the terms: $9x - 5x = 2458.35 - 160$.$4x = 2298.35$.$x = \frac{2298.35}{4} = 574.5875$.Rounding to the nearest provided option,we get $x = 574.25$ (using $273$ as the conversion constant).Using $K = 273$: $\frac{x - 32}{9} = \frac{x - 273}{5} \implies 5x - 160 = 9x - 2457 \implies 4x = 2297 \implies x = 574.25$.

The temperature of a body on the Kelvin scale is found to be $x \ K$. When it is measured by a Fahrenheit thermometer,it is found to be $x \ ^\circ F$. The value of $x$ is:

Similar Questions

VedClass-Data Entry

What is the structure and hybridization of $Si(CH_3)_4$?

Sulphur on boiling with $NaOH$ solution gives

$A$ given charge is situated at a certain distance from an electric dipole in the end-on position experiences a force $F$. If the distance of the charge is doubled,the force acting on the charge will be

$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3\, cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion?.......$cm$

Vedclass Test Series

Exam Paper Generator

Online Exam Module