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 temperature in Celsius $(T_C)$ and Fahrenheit $(T_F)$ is given by $T_F = \frac{9}{5} T_C + 32$.We know that $T_C = T_K -

Vedclass · Accepted Answer

$574.25$

Vedclass · Answer

(C) The relationship between temperature in Celsius $(T_C)$ and Fahrenheit $(T_F)$ is given by $T_F = \frac{9}{5} T_C + 32$.We know that $T_C = T_K - 273.15$,where $T_K$ is the temperature in Kelvin.Substituting $T_K = x$ and $T_F = x$,we get:$x = \frac{9}{5}(x - 273.15) + 32$$x = 1.8(x - 273.15) + 32$$x = 1.8x - 491.67 + 32$$x = 1.8x - 459.67$$0.8x = 459.67$$x = \frac{459.67}{0.8} = 574.5875 \approx 574.25$ (using standard approximation $273$ instead of $273.15$ gives $x = 574.25$ exactly).Thus,$x = 574.25$.

The temperature of a body on the Kelvin scale is $x \ K$. When the temperature of this body is measured with a Fahrenheit thermometer,it is found to be $x \ ^\circ F$. Then $x = $......

Similar Questions

What is absolute zero temperature?

If the reading in Fahrenheit scale is twice the reading in Celsius scale, then the reading in Fahrenheit scale is (in $^{\circ} F$)

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

On a new scale of temperature (which is linear) called the $W$ scale,the freezing and boiling points of water are $39\,^{\circ}W$ and $239\,^{\circ}W$ respectively. What will be the temperature on the new scale,corresponding to a temperature of $39\,^{\circ}C$ on the Celsius scale?

$A$ thermoelectric thermometer is based on:

Vedclass Test Series

Exam Paper Generator

Online Exam Module