
The temperature on Celsius scale is $25^\circ {\rm{C}}$. What is the
corresponding temperature on the Fahrenheit scale
A. $40^\circ {\rm{F}}$
B. $77^\circ {\rm{F}}$
C. $50^\circ {\rm{F}}$
D. $45^\circ {\rm{F}}$
Answer
578.7k+ views
Hint: The degree centigrade and Fahrenheit are the units to measure the temperature. In the solution we will use the relation between the temperature on Celsius scale and temperature on the Fahrenheit scale to convert the $^\circ {\rm{C}}$ into \[^\circ{\rm{F}}\].
Complete step by step answer:
Given:
The on Celsius scale is $25^\circ {\rm{C}}$.
We need to find the temperature on Fahrenheit scale.
The temperature range in Celsius was first defined by setting the temperature at zero at which water froze. This temperature was redefined as the temperature at which ice melts. The other point at which Celsius was set is $100^\circ {\rm{C}}$ which is defined as the boiling point of water.
This Celsius range is also defined in Kelvin. Zero degrees Celsius is now defined as $273.15\;{\rm{K}}$. So, the boiling point of the water will be equal to the $273.15 + 100 =373.15\;{\rm{K}}$.
For Fahrenheit temperature, the range is based on setting the freezing point of water at $32$ degrees, and boiling point at $212$ degrees. Absolute zero is defined as $ -459.67^\circ {\rm{F}}$.
The relation between temperature on Celsius scale and temperature on the Fahrenheit scale
is given by,
${T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}{T_{\left( {^\circ {\rm{C}}} \right)}} + 32$
Here, ${T_{\left( {^\circ {\rm{C}}} \right)}}$ is the temperature in Celsius scale, and ${T_{\left( {^\circ {\rm{F}}} \right)}}$ is the temperature in Fahrenheit scale.
Find the temperature on Fahrenheit scale.
Substituting $25^\circ {\rm{C}}$ for ${T_{\left( {^\circ {\rm{C}}} \right)}}$ in the equation
${T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}{T_{\left( {^\circ {\rm{C}}} \right)}} + 32$ to
get the value of temperature in Fahrenheit scale.
$
{T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}\left( {25} \right) + 32\\
= 45 + 32\\
= 77^\circ {\rm{F}}
$
So, the correct answer is “Option B”.
Note:
Make sure to use positive signs between the relationship because, and perform the correct calculation. Sometimes students make mistakes by taking the $\dfrac{5}{9}$ in place of $\dfrac{9}{5}$ which results in the wrong answer.
Complete step by step answer:
Given:
The on Celsius scale is $25^\circ {\rm{C}}$.
We need to find the temperature on Fahrenheit scale.
The temperature range in Celsius was first defined by setting the temperature at zero at which water froze. This temperature was redefined as the temperature at which ice melts. The other point at which Celsius was set is $100^\circ {\rm{C}}$ which is defined as the boiling point of water.
This Celsius range is also defined in Kelvin. Zero degrees Celsius is now defined as $273.15\;{\rm{K}}$. So, the boiling point of the water will be equal to the $273.15 + 100 =373.15\;{\rm{K}}$.
For Fahrenheit temperature, the range is based on setting the freezing point of water at $32$ degrees, and boiling point at $212$ degrees. Absolute zero is defined as $ -459.67^\circ {\rm{F}}$.
The relation between temperature on Celsius scale and temperature on the Fahrenheit scale
is given by,
${T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}{T_{\left( {^\circ {\rm{C}}} \right)}} + 32$
Here, ${T_{\left( {^\circ {\rm{C}}} \right)}}$ is the temperature in Celsius scale, and ${T_{\left( {^\circ {\rm{F}}} \right)}}$ is the temperature in Fahrenheit scale.
Find the temperature on Fahrenheit scale.
Substituting $25^\circ {\rm{C}}$ for ${T_{\left( {^\circ {\rm{C}}} \right)}}$ in the equation
${T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}{T_{\left( {^\circ {\rm{C}}} \right)}} + 32$ to
get the value of temperature in Fahrenheit scale.
$
{T_{\left( {^\circ {\rm{F}}} \right)}} = \dfrac{9}{5}\left( {25} \right) + 32\\
= 45 + 32\\
= 77^\circ {\rm{F}}
$
So, the correct answer is “Option B”.
Note:
Make sure to use positive signs between the relationship because, and perform the correct calculation. Sometimes students make mistakes by taking the $\dfrac{5}{9}$ in place of $\dfrac{9}{5}$ which results in the wrong answer.
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