Chapter 1, Problem 1.111QP

In January 2009, the National Aeronautics and Space Administration (NASA) reported that a planet in our galaxy, known as HD 80606b, underwent a temperature change from 980°F to 2240°F over the course of six hours, (a) Convert these temperatures and the range they span to degrees Celsius, and to kelvins. (b) Determine the rate of temperature change per second in degrees Fahrenheit, degrees Celsius, and kelvins.

Interpretation Introduction

Interpretation: From the given details,

It should be converted the given temperature range to degree Celsius and to Kelvin.

It should be determined the rate of temperature change per second in degrees Fahrenheit, degree Celsius and Kelvins

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Fahrenheit to Celsius

  ${}^{\text{o}}\text{C}=({\text{F}}^{\text{o}}-{\text{32F}}^{\text{o}})\times \frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}$

• Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

• $\text{Rate}\text{of}\text{temperature}\text{change}\text{per}\text{second}\text{=}\frac{\text{Change}\text{in}\text{temperature}}{\text{number}\text{of}\text{seconds}}$

Answer to Problem 1.111QP

• Temperature range in degree Celsius is $\text{527°C}\text{to}\text{1227°C}$
• Temperature range in Kelvin is $\text{800K}\text{to}\text{1500K}$
• Rate of temperature change per second in degrees Fahrenheit is $\text{5}\text{.833}{\text{×10}}^{\text{-2}}\text{°F/s}$
• Rate of temperature change per second in degree Celsius is $\text{3}\text{.241}{\text{×10}}^{\text{-2}}\text{°C/s}$
• Rate of temperature change per second in Kelvin is $\text{3}\text{.241}{\text{×10}}^{\text{-2}}\text{K/s}$

Explanation of Solution

The given temperature range can be converted to degree Celsius and to Kelvin

 By using the conversion formula, that is

Conversional formula from Fahrenheit to Celsius

${}^{\text{o}}\text{C}=({\text{F}}^{\text{o}}-{\text{32F}}^{\text{o}})\times \frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}$

Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

Here the temperature range in Fahrenheit is $\text{980}°\text{F}\text{to}\text{2240}°\text{F}$

Conversion from Fahrenheit to Celsius

Firstly it is needed to convert $\text{980}°\text{F}$ to degree Celsius

$\begin{array}{l}{}^{\text{o}}\text{C=}{\text{(F}}^{\text{o}}{\text{-32F}}^{\text{o}}\text{)}\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {}^{\text{o}}\text{C=}{\text{(980F}}^{\text{o}}{\text{-32F}}^{\text{o}}\text{)}\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\text{=}\text{527°C}\end{array}$

Second conversion is $\text{2240}°\text{F}$ to degree Celsius

$\begin{array}{l}{}^{\text{o}}\text{C=}{\text{(F}}^{\text{o}}{\text{-32F}}^{\text{o}}\text{)}\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {}^{\text{o}}\text{C=}{\text{(2240F}}^{\text{o}}{\text{-32F}}^{\text{o}}\text{)}\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\text{=}122\text{7°C}\end{array}$

The given temperature range in degree Celsius is $\text{527°C}\text{to}\text{1227°C}$

Conversion from Celsius to Kelvin

The given temperature range in degree Celsius is $\text{527°C}\text{to}\text{1227°C}$

Conversional formula from Celsius to Kelvin is

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

Firstly it is needed to convert  $527\text{°C}$ to Kelvin

$\text{K}\text{=}{527}^{\circ }\text{C}\text{+}\text{273}\text{.15}\text{=}\text{800K}$

Secondly it is needed to convert  $122\text{7°C}$ to Kelvin

$\text{K}\text{=}{1227}^{\circ }\text{C}\text{+}\text{273}\text{.15}\text{=}1500\text{K}$

The given temperature range in Kelvin is $\text{800K}\text{to}\text{1500K}$

The rate of temperature change per second in degrees Fahrenheit, degree Celsius and Kelvins

$\text{Rate}\text{of}\text{temperature}\text{change}\text{per}\text{second}\text{=}\frac{\text{Change}\text{in}\text{temperature}}{\text{number}\text{of}\text{seconds}}$

Rate of temperature change per second in degrees Fahrenheit can be determined as follows;

$\begin{array}{l}\text{2240°F}\text{-980°F}\text{=}\text{1260°F}\\ \frac{\text{1260°F}}{\text{6h}}\text{×}\frac{\text{1h}}{\text{60}\text{min}}\text{×}\frac{\text{1}\text{min}}{\text{60}\text{s}}\text{=}\text{5}\text{.833}{\text{×10}}^{\text{-2}}\text{°F/s}\end{array}$

 Rate of temperature change per second in degrees Celsius can be determined as follows;

$\begin{array}{l}\text{1227°C}\text{-527°C}\text{=}700\text{°C}\\ \frac{\text{700°C}}{\text{6h}}\text{×}\frac{\text{1h}}{\text{60}\text{min}}\text{×}\frac{\text{1}\text{min}}{\text{60}\text{s}}\text{=}3.241{\text{×10}}^{\text{-2}}\text{°C/s}\end{array}$

Rate of temperature change per second in Kelvin can be determined as follows;

$\begin{array}{l}1500K\text{-800K}\text{=}700K\\ \frac{\text{700K}}{\text{6h}}\text{×}\frac{\text{1h}}{\text{60}\text{min}}\text{×}\frac{\text{1}\text{min}}{\text{60}\text{s}}\text{=}3.241{\text{×10}}^{\text{-2}}\text{°K/s}\end{array}$

Conclusion

The rate of temperature change per second in degrees Fahrenheit, degree Celsius and Kelvins is determined.