
A thermometer graduated according to a linear scale reads a value ${x_0}$ when in contact with boiling water, and $\dfrac{{{x_0}}}{3}$ when in contact with ice. What is the temperature of an object in ${0^ \circ }C$, if this thermometer in contact with the object reads $\dfrac{{{x_0}}}{2}$?
A. 35
B. 25
C. 60
D. 40
Answer
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Hint: To calculate this problem, we need to use the values of standard temperatures of boiling point and the freezing point of water in the Celsius scale, which serve as the basis for calibrating the temperature scale.
In the Celsius scale, the temperature corresponding to the freezing point of water is ${0^ \circ }C$ and the temperature corresponding to the boiling point of water is ${100^ \circ }C$.
Complete step by step answer:
The temperature is different from heat. The heat is defined as the energy which is transferred across the system which can change the net total of the internal energy of the system whereas, the temperature is a quantity that manifests out of the heat energy, that measures the degree to decide the direction and the extent of transfer of heat energy from one system to another.
The temperature is measured by a device known as a thermometer. It is a device which contains a liquid such as alcohol or mercury is a small bulb, which expands into a small vertical capillary column, which has several graduation markings depending on the scale used to measure the temperature.
The graduations which are marked on the thermometer is based on a standard scale of measurement. Historically, there have been several scales of temperature, but the most common and the widely used scales of temperature are only 3: i) Celsius scale ii) Fahrenheit scale iii) Kelvin scale
Let us understand in detail about the Celsius scale:
Also known as the centigrade scale, the Celsius scale consists of 100 divisions starting from ${0^ \circ }C$ which is the first reference point called the freezing point of water and the highest is the second reference point, the boiling point of water which is ${100^ \circ }C$. This scale is widely used in almost all countries except the USA, for commercial purposes like meteorology.
Hence, any thermometer calibrated to the Celsius scale should have two fixed points of reference, mainly the freezing point of water at ${0^ \circ }C$ and boiling point of water at ${100^ \circ }C$
Given in the problem,
Marking of the second reference point of the scale at ${100^ \circ }C$ = ${x_0}$
Marking of the first reference point of the scale at ${0^ \circ }C$ = $\dfrac{{{x_0}}}{3}$
The temperature difference with respect to the scale,
${x_0} - \dfrac{{{x_0}}}{3} = 100 - 0$
$ \Rightarrow \dfrac{{3{x_0} - {x_0}}}{3} = 100$
$ \Rightarrow 2{x_0} = 300$
$ \Rightarrow {x_0} = 150$
Thus, the value of ${x_0} = 150$
The reading of the temperature of the object is $\dfrac{{{x_0}}}{2}$. However, this is the absolute value, but to calculate the actual value of the temperature, we have to take the reference of the melting point of ice, as measured by this scale i.e. $\dfrac{{{x_0}}}{3}$.
The temperature of the object with respect to the melting point of the ice as taken by this scale is,
$T = \dfrac{{{x_0}}}{2} - \dfrac{{{x_0}}}{3} = \dfrac{{{x_0}}}{6}$
Substituting the value of ${x_0} = 150$, we have –
$T = \dfrac{{150}}{6} = {25^ \circ }C$
Hence, the temperature of the object is ${25^ \circ }C$
Hence, the correct option is Option B.
Note: The terms melting point of ice and freezing point of water are the same in the Celsius scale at ${0^ \circ }C$. This usage of these terms can be easily interchanged in any question. The students should take heed of the interchangeability of these terms.
Complete step by step answer:
The temperature is different from heat. The heat is defined as the energy which is transferred across the system which can change the net total of the internal energy of the system whereas, the temperature is a quantity that manifests out of the heat energy, that measures the degree to decide the direction and the extent of transfer of heat energy from one system to another.
The temperature is measured by a device known as a thermometer. It is a device which contains a liquid such as alcohol or mercury is a small bulb, which expands into a small vertical capillary column, which has several graduation markings depending on the scale used to measure the temperature.
The graduations which are marked on the thermometer is based on a standard scale of measurement. Historically, there have been several scales of temperature, but the most common and the widely used scales of temperature are only 3: i) Celsius scale ii) Fahrenheit scale iii) Kelvin scale
Let us understand in detail about the Celsius scale:
Also known as the centigrade scale, the Celsius scale consists of 100 divisions starting from ${0^ \circ }C$ which is the first reference point called the freezing point of water and the highest is the second reference point, the boiling point of water which is ${100^ \circ }C$. This scale is widely used in almost all countries except the USA, for commercial purposes like meteorology.
Hence, any thermometer calibrated to the Celsius scale should have two fixed points of reference, mainly the freezing point of water at ${0^ \circ }C$ and boiling point of water at ${100^ \circ }C$
Given in the problem,
Marking of the second reference point of the scale at ${100^ \circ }C$ = ${x_0}$
Marking of the first reference point of the scale at ${0^ \circ }C$ = $\dfrac{{{x_0}}}{3}$
The temperature difference with respect to the scale,
${x_0} - \dfrac{{{x_0}}}{3} = 100 - 0$
$ \Rightarrow \dfrac{{3{x_0} - {x_0}}}{3} = 100$
$ \Rightarrow 2{x_0} = 300$
$ \Rightarrow {x_0} = 150$
Thus, the value of ${x_0} = 150$
The reading of the temperature of the object is $\dfrac{{{x_0}}}{2}$. However, this is the absolute value, but to calculate the actual value of the temperature, we have to take the reference of the melting point of ice, as measured by this scale i.e. $\dfrac{{{x_0}}}{3}$.
The temperature of the object with respect to the melting point of the ice as taken by this scale is,
$T = \dfrac{{{x_0}}}{2} - \dfrac{{{x_0}}}{3} = \dfrac{{{x_0}}}{6}$
Substituting the value of ${x_0} = 150$, we have –
$T = \dfrac{{150}}{6} = {25^ \circ }C$
Hence, the temperature of the object is ${25^ \circ }C$
Hence, the correct option is Option B.
Note: The terms melting point of ice and freezing point of water are the same in the Celsius scale at ${0^ \circ }C$. This usage of these terms can be easily interchanged in any question. The students should take heed of the interchangeability of these terms.
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