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Q4PE

Expert-verifiedFound in: Page 470

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit temperature? What is this on the Celsius scale?**

We get the corresponding temperature in Fahrenheit is ${4760}{\xb0}{\mathrm{F}}$and in Celcius is 2600℃.

**We can find Fahrenheit temperature from Kelvin temperature by the given equation,**

${\mathbf{T}}{\left(\xb0F\right)}{\mathbf{=}}\frac{\mathbf{9}}{\mathbf{5}}{\left[T\left(K\right)-273.15\right]}{\mathbf{+}}{\mathbf{32}}$

The temperature in Kelvin is given,

${\mathbf{T}}{\left(K\right)}{\mathbf{=}}{\mathbf{2900}}{\mathbf{K}}$

Adding this to the equation,

${\mathbf{T}}{\left(\xb0F\right)}{\mathbf{=}}\frac{\mathbf{9}}{\mathbf{5}}{\left(2900-273.15\right)}{\mathbf{+}}{\mathbf{32}}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}{\left(\xb0F\right)}{\mathbf{=}}\frac{\mathbf{9}}{\mathbf{5}}{\mathbf{\times}}{\mathbf{2626}}{\mathbf{.}}{\mathbf{85}}{\mathbf{+}}{\mathbf{32}}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}{\left(\xb0F\right)}{\mathbf{=}}{\mathbf{4728}}{\mathbf{.}}{\mathbf{33}}{\mathbf{+}}{\mathbf{32}}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}{\left(\xb0F\right)}{\mathbf{=}}{\mathbf{4760}}{\mathbf{.}}{\mathbf{33}}$

So. ${\mathbf{4760}}{\mathbf{.}}{\mathbf{33}}{\mathbf{\xb0}}{\mathbf{F}}$ is the Fahrenheit temperature equal to the Kelvin scale temperature of 2900 K.

Kelvin scale related to Celcius scale temperature through the equation,

$\begin{array}{rcl}{\mathbf{T}}{\left(\xb0C\right)}& {\mathbf{=}}& {\mathbf{T}}{\left(K\right)}{\mathbf{-}}{\mathbf{273}}{\mathbf{.}}{\mathbf{15}}\\ {\mathbf{T}}{\left(\xb0C\right)}& {\mathbf{=}}& {\mathbf{2900}}{\mathbf{-}}{\mathbf{273}}{\mathbf{.}}{\mathbf{15}}\\ {\mathbf{T}}{\left(\xb0C\right)}& {\mathbf{=}}& {\mathbf{2626}}{\mathbf{.}}{\mathbf{85}}\\ & & \end{array}$

So, the temperature on the Celcius scale is, or approximately equal to ${\mathbf{2600}}{\mathbf{\xb0}}{\mathit{C}}$

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