Q43P - Chemistry: Molecular Nature Of Matter And Change

Question

Perform the following conversions: 

(a) 106°F (the body temperature of many birds) to K and °C 

(b) 3410°C (the melting point of tungsten, the highest for any metallic element) to K and °F 

(c) 6.1ᵡ10³ K (the surface temperature of the Sun) to °F and °C

Step-by-Step Solution

Answer

There are three main scales for measuring temperature, these are Kelvin scale, Celsius scale and Fahrenheit scale. The values of temperature given in one scale can be converted to another by keeping in mind various factors in which they differ like size of the unit and zero point. Using these in mind the answers of the given questions are

1.   ${106}^{\circ }F=41.{11}^{\circ }C=314.26K$
2.  ${3410}^{\circ }C={6170}^{\circ }F=3683.15K$
3. $6.1\times {10}^{3}K=5.8\times {{10}^3}^{\circ }C=10.52\times {{10}^3}^{\circ }F$
   

Step 1: Scales of Temperature and their Interconversion

Relation between Celsius and Fahrenheit scales:

$\mathit{T}\mathbf{\left(}\mathit{i}{\mathit{n}}^{\mathbf{\circ }}\mathit{F}\mathbf{\right)}\mathbf{=}\frac{\mathbf{9}}{\mathbf{5}}\mathit{T}\mathbf{\left(}\mathit{i}{\mathit{n}}^{\mathbf{\circ }}\mathit{C}\mathbf{\right)}\mathbf{+}\mathbf{32}$

Relation between Celsius and Kelvin scales: 

$\mathit{T}\mathbf{\left(}\mathit{i}\mathit{n}\mathit{K}\mathbf{\right)}\mathbf{=}\mathit{T}\mathbf{\left(}\mathit{i}{\mathit{n}}^{\mathbf{\circ }}\mathit{C}\mathbf{\right)}\mathbf{+}\mathbf{273}\mathbf{.}\mathbf{15}$

Step 2: 106°F (the body temperature of many birds) to K and °C

First, we need to convert °F to °C by using the relation,

$\mathit{T}\mathbf{\left(}\mathit{i}{\mathit{n}}^{\mathbf{\circ }}\mathit{F}\mathbf{\right)}\mathbf{=}\frac{\mathbf{9}}{\mathbf{5}}\mathit{T}\mathbf{\left(}\mathit{i}{\mathit{n}}^{\mathbf{\circ }}\mathit{C}\mathbf{\right)}\mathbf{+}\mathbf{32}$

So, $T{(}^{\circ }C)=\frac{5}{9}(T i{n}^{\circ }F-32)=\frac{5}{9}(106-32)=41.{11}^{\circ }C$

And then °C to K,

T(in K) = T(in ^oC) + 273.15 = 41.11 + 273.15 = 314.26K

Hence, 106 ^oF = 41.11 ^oC = 314.26 K

Step 3: 3410°C (the melting point of tungsten, the highest for any metallic element) to K and °F

Converting Celsius to Kelvin using the formula given in step 1

T(in^oF) = T(in^oC) + 273.15 = 3410 + 273.15 = 3683.15 K

Converting Celsius to Fahrenheit

$T\left(i{n}^{\circ }F\right)=\frac{9}{5}T\left(i{n}^{\circ }C\right)+32=\frac{9}{5}\left(3410\right)+32={6170}^{\circ }F$

Hence, 3410^oC = 6170^oF = 3683.15 K

Step 4: 6.1ᵡ10 3 K (the surface temperature of the Sun) to °F and °C

First convert Kelvin to Celsius using the formula given in step 1

T(in^oC) = T(in K) - 273.15 = 6100 - 273.15 = 5826.85^oC

Now convert Celsius to Fahrenheit

$T\left(i{n}^{\circ }F\right)=\frac{9}{5}T\left(i{n}^{\circ }C\right)+32=\frac{9}{5}(5826.85)+32=10.52\times {{10}^3}^{\circ }F$

Hence, $6.1\times {10}^{3}K=5.8\times {{10}^3}^{\circ }C=10.52\times {{10}^3}^{\circ }F$