Express 98^oF into Kelvin scale.

Solution

F = 98^oF

K =?

We know that,

[latex]\frac{F-32}{180} = \frac{K-273}{100}[/latex]

Or, [latex]\frac{98-32}{18} = \frac{K-273}{10}[/latex]

Or, [latex]\frac{66}{18} = \frac{K-273}{10}[/latex]

Or, 660 = 18K – 4914

Or, 18K = 660 + 4914

Or, 18K = 5574

Or, K = 309.67 K

At what temperature will the Kelvin scale reading double the Fahrenheit reading?

Solution

Let, Fahrenheit reading (F) = x

So, Kelvin reading (K) = 2x

We know that,

[latex]\frac{F-32}{180} = \frac{K-273}{100}[/latex]

Or, [latex]\frac{x-32}{180} = \frac{2x-273}{100}[/latex]

Or, [latex]\frac{x-32}{9} = \frac{2x-273}{5}[/latex]

Or, 18x – 2457 = 5x – 160

Or, 13x = 2297

Or, x = 176.69            

So, required answer is: 176.69^oF or 353.38K

At what temperature, do the Fahrenheit-thermometer and Celsius-thermometer show the same reading?

Solution

Let, Fahrenheit reading (F) = x

So, Celsius reading (C) = x

We know that,

[latex]\frac{F-32}{180} = \frac{C}{100}[/latex]

Or, [latex]\frac{x-32}{180} = \frac{x}{100}[/latex]

Or, [latex]\frac{x-32}{9} = \frac{x}{5}[/latex]

Or, 5x – 160 = 9x

Or, – 4x = 160            

Or, x = – 40^oC

At what point of thermometric scale does Kelvin scale reading coincide with Fahrenheit scale reading?

Solution

Point where Kelvin and Fahrenheit scale reading coincides =?

i.e., K = F = x let,

Using formula,

[latex]\frac{F-32}{180} = \frac{F-273}{100}[/latex]

Or, [latex]\frac{x-32}{18} = \frac{x-273}{10}[/latex]

Or, 10x – 320 = 18x – 4914

Or, – 320 + 4914 = 18x – 10x

Or, 4594 = 8x

Or, x = 574.25K

A centigrade thermometer reads 1^oC at the melting point of ice and 99^oC at the boiling point of water at normal pressure. What is the correct temperature when it reads 25^oC? Ans: 24.5^oC

Solution:

Faulty Thermometer

LFP = 1^oC

UFP = 99^oC

Reading shown = 25^oC

Correct Thermometer

LFP = 0^oC

UFP = 100^oC

Reading shown = C

We know,

[latex]\frac{X-L}{U-L} = \frac{C-0}{100-0}[/latex]

Or, [latex]\frac{25-1}{99-1} = \frac{C}{100}[/latex]

Or, [latex]\frac{24}{98} = \frac{C}{100}[/latex]

Or, C = [latex]\frac{2400}{98} = 24.49^o\ C[/latex]

A faulty Fahrenheit thermometer shows a temperature of 206^oF when immersed in boiling water. When a correct Celsius thermometer is dipped in the same water it records 96^oC. Find the correction to be applied to the faulty thermometer.

Solution

Correct temperature on Celsius thermometer (C) = 96^oC

We know,

[latex]\frac{C}{100} = \frac{F-32}{180}[/latex]

Or, [latex]\frac{96}{5} = \frac{F-32}{9}[/latex]

Or, 5F – 160 = 864

Or, F = 204.8^oF

So, correct temperature on Fahrenheit scale = 204.8^oF              

[latex]\therefore[/latex] Correction to be applied = 206 – 204.8 = 1.2^oF

What is the change in Fahrenheit scale when the temperature change in Centigrade scale is 12.5?

Solution

Change in Fahrenheit scale [latex](\Delta F)[/latex] =?

Change in centigrade scale [latex](\Delta C)[/latex] =?

We have,

[latex]\frac{F-32}{180} = \frac{C}{100}[/latex]

Or, [latex]\frac{F-32}{9} = \frac{C}{5}[/latex]

Or, C = [latex]\frac{5}{9}(F-32)[/latex]

So, [latex]\Delta C = \frac{5}{9}(\Delta F)[/latex]           

Or, [latex]\Delta F = \frac{9}{5}\times 12.5 = 22.5^o F[/latex]