Temperature and Temperature scales Questions in English

(A) The relationship between the Celsius scale and the Kelvin scale is given by the formula: $T(K) = T(^{\circ}C) + 273.15$.
Substituting $T(^{\circ}C) = 0^{\circ}C$ into the formula:
$T(K) = 0 + 273.15 = 273.15 \ K$.
Therefore,the correct value is $273.15 \ K$.

(D) The relationship between the Kelvin scale $(T)$ and the Celsius scale $(t)$ is given by the formula: $T = t + 273.15$.
Absolute zero is defined as $0 \ K$ on the Kelvin scale.
Substituting $T = 0$ into the equation: $0 = t + 273.15$.
Solving for $t$,we get: $t = -273.15 \ ^\circ C$.
Therefore,the absolute zero of temperature on the Celsius scale is $-273.15 \ ^\circ C$.

(B) The relationship between the Celsius scale $(C)$ and the Fahrenheit scale $(F)$ is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
Given $C = -183^{\circ}C$.
Substituting the value into the formula: $\frac{-183}{5} = \frac{F - 32}{9}$.
$-36.6 = \frac{F - 32}{9}$.
$-36.6 \times 9 = F - 32$.
$-329.4 = F - 32$.
$F = -329.4 + 32 = -297.4^{\circ}F$.
Rounding to the nearest integer,we get $F \approx -297^{\circ}F$.

(A) The relationship between the Kelvin scale $(K)$ and the Fahrenheit scale $(F)$ is given by the formula: $\frac{F - 32}{9} = \frac{K - 273}{5}$.
Substituting the given value $K = 95\, K$ into the equation:
$\frac{F - 32}{9} = \frac{95 - 273}{5}$
$\frac{F - 32}{9} = \frac{-178}{5}$
$\frac{F - 32}{9} = -35.6$
$F - 32 = -35.6 \times 9$
$F - 32 = -320.4$
$F = -320.4 + 32$
$F = -288.4^\circ F$.
Rounding to the nearest integer,we get $F \approx -288^\circ F$.

(C) The relationship between the Celsius scale and the Kelvin scale is given by $T(K) = T(^{\circ}C) + 273.15$.
For a change in temperature,$\Delta T(K) = \Delta T(^{\circ}C)$.
Since the increase in temperature is $27^{\circ}C$,the corresponding increase on the Kelvin scale is $27\, K$.

(D) platinum resistance thermometer is highly accurate and stable over a wide range of temperatures. It is specifically designed to measure temperatures ranging from $-200^{\circ}C$ to $600^{\circ}C$. Due to the linear relationship between the resistance of platinum and temperature,it is considered the most suitable instrument for this specific range.

(B) thermoelectric thermometer operates on the principle of the Seebeck effect.
In the Seebeck effect,a temperature difference between two dissimilar electrical conductors or semiconductors produces a voltage difference between them.
This phenomenon is utilized in thermocouples to measure temperature by converting thermal energy directly into electrical energy.

(B) The maximum density of water occurs at $4^\circ C$.
To convert this temperature to Fahrenheit, we use the formula:
$\frac{C}{5} = \frac{F - 32}{9}$
Substituting $C = 4$ into the equation:
$\frac{4}{5} = \frac{F - 32}{9}$
$0.8 = \frac{F - 32}{9}$
$7.2 = F - 32$
$F = 39.2^\circ F$
Therefore, the correct option is $B$.

(C) The branch of physics that deals with the production and behavior of materials at very low temperatures,typically below $123\,K$ (the temperature of liquid nitrogen),is known as Cryogenics.
It involves the study of physical phenomena at temperatures close to $0\,K$ (absolute zero).

(C) The absolute scale of temperature (Kelvin scale) is defined based on the properties of an ideal gas.
In practical laboratory conditions,a constant volume helium gas thermometer is used to reproduce this scale because helium behaves very closely to an ideal gas over a wide range of temperatures.
Therefore,the correct option is $C$.

(D) The correct option is $D$.
The absolute zero temperature is $0 \ K$,which corresponds to $-273.15^\circ C$.
Since the Kelvin scale starts from absolute zero $(0 \ K)$,which is the lowest possible theoretical temperature,the temperature on the Kelvin scale can never be negative.

(B) The relationship between the Celsius scale $(C)$ and the Fahrenheit scale $(F)$ is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
Given $C = 25^{\circ}C$,substitute this value into the equation:
$\frac{25}{5} = \frac{F - 32}{9}$
$5 = \frac{F - 32}{9}$
$45 = F - 32$
$F = 45 + 32 = 77^{\circ}F$.
Therefore,the correct option is $B$.

(C) When a thermometer bulb is wrapped in a wet hanky,the water evaporates from the surface of the hanky.
Evaporation is a cooling process that absorbs latent heat from the thermometer bulb.
As a result,the temperature of the wet-bulb thermometer decreases.
Therefore,the dry-bulb thermometer will record a higher temperature than the wet-bulb thermometer.

(A) The relationship between the Celsius scale $(C)$ and the Fahrenheit scale $(F)$ is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
To find the temperature where both readings are the same,let $C = F = t$.
Substituting this into the formula,we get: $\frac{t}{5} = \frac{t - 32}{9}$.
Cross-multiplying gives: $9t = 5(t - 32)$.
$9t = 5t - 160$.
$9t - 5t = -160$.
$4t = -160$.
$t = -40$.
Therefore,at $-40^\circ$,the readings on both the Celsius and Fahrenheit scales are identical.

(D) The standardisation of thermometers is performed using a gas thermometer.
Gas thermometers,particularly constant-volume gas thermometers,are considered the primary standard for temperature measurement because the properties of gases (like pressure and volume) follow the ideal gas law $PV = nRT$ very closely,providing a highly accurate and reproducible scale across a wide range of temperatures.

(A) The sensitivity of a thermometer depends on the coefficient of volume expansion $(\gamma)$.
For gases,the coefficient of volume expansion is significantly higher than that of liquids.
Since $\Delta V = V_0 \gamma \Delta T$,a larger value of $\gamma$ results in a larger change in volume for a given change in temperature $(\Delta T)$.
Therefore,gas thermometers are more sensitive than liquid thermometers.

(C) The boiling point of mercury is approximately $357^oC$.
$A$ mercury thermometer operates on the principle of thermal expansion of liquid mercury with a rise in temperature.
It is typically used to measure temperatures in the range of $-30^oC$ to $357^oC$.
Therefore,among the given options,the mercury thermometer can be used to measure temperatures up to $360^oC$ (as it is the closest practical upper limit provided).

(B) The boiling point of a liquid increases with an increase in external pressure. By filling the space above the mercury column with an inert gas like nitrogen at high pressure,the boiling point of mercury is raised significantly above its normal boiling point of $367^{\circ}C$. This allows the thermometer to measure temperatures up to $500^{\circ}C$ without the mercury boiling.

(A) $Pyrometer$ is a type of remote-sensing thermometer used to measure the temperature of a surface.
It is specifically designed to measure very high temperatures,such as those found in furnaces or industrial processes,by detecting the thermal radiation emitted by the object.
Therefore,the correct option is $A$.

(C) The relationship between the Fahrenheit scale $(F)$ and the Kelvin scale $(K)$ is given by the formula: $\frac{F - 32}{9} = \frac{K - 273.15}{5}$.
Absolute zero is defined as $0 \ K$.
Substituting $K = 0$ into the equation:
$\frac{F - 32}{9} = \frac{0 - 273.15}{5}$
$\frac{F - 32}{9} = -54.63$
$F - 32 = -54.63 \times 9$
$F - 32 = -491.67$
$F = -491.67 + 32 = -459.67^\circ F$.
Rounding to the nearest whole number,we get $-460^\circ F$.

(C) According to Charles's law, for a gas at constant pressure, the volume $V$ is directly proportional to the absolute temperature $T$ ($V \propto T$ or $\frac{V}{T} = \text{constant}$).

Given:
Initial volume $V_1 = 47.5$ units at temperature $T_1 = 0^\circ C = 273 \, K$.
Final volume $V_2 = 67$ units at boiling point temperature $T_2$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$T_2 = \frac{V_2 \times T_1}{V_1}$
$T_2 = \frac{67 \times 273}{47.5} \approx 385 \, K$.
To convert the temperature from Kelvin to Celsius:
$t(^\circ C) = T(K) - 273$
$t = 385 - 273 = 112^\circ C$.
Thus, the boiling point of the liquid is $112^\circ C$.

(A) The temperature on any scale can be converted to another scale using the linear relationship:
$\frac{X - LFP}{UFP - LFP} = \text{constant}$
Where $X$ is the temperature reading,$LFP$ is the Lower Fixed Point (freezing point),and $UFP$ is the Upper Fixed Point (boiling point).
For the Celsius scale: $LFP = 0^{\circ}C$,$UFP = 100^{\circ}C$.
For the given thermometer: $LFP' = 20^{\circ}C$,$UFP' = 150^{\circ}C$.
Equating the two:
$\frac{X - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\frac{X - 20}{130} = \frac{60}{100}$
$\frac{X - 20}{130} = 0.6$
$X - 20 = 0.6 \times 130$
$X - 20 = 78$
$X = 98^{\circ}C$.

(D) The relationship between Celsius $(C)$ and Fahrenheit $(F)$ scales is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
Given $F = 140^{\circ}F$.
Substituting the value of $F$ in the formula:
$\frac{C}{5} = \frac{140 - 32}{9}$
$\frac{C}{5} = \frac{108}{9}$
$\frac{C}{5} = 12$
$C = 12 \times 5 = 60^{\circ}C$.
Therefore,the correct option is $D$.

(A) thermocouple thermometer consists of two dissimilar metal wires joined at two junctions.
It works on the Seebeck effect, where a temperature difference between the junctions produces an electromotive force $(EMF)$.
Because thermocouples have a very low thermal mass and high thermal conductivity, they respond very quickly to changes in temperature.
Therefore, they are the ideal choice for measuring rapidly changing temperatures compared to gas, resistance, or vapour pressure thermometers, which have higher thermal inertia.

(A) The relationship between the Celsius scale and the Kelvin scale is given by the formula: $T(K) = T(^{\circ}C) + 273.15$.
Substituting the value of $0^{\circ}C$ into the formula:
$T(K) = 0 + 273.15 = 273.15 \ K$.
Therefore,$0^{\circ}C$ is equivalent to $273.15 \ K$ on the Kelvin scale.

(D) The density of water is anomalous. As water is cooled from room temperature,its density increases until it reaches $4^{\circ}C$. At $4^{\circ}C$,water molecules are packed most closely together,resulting in maximum density. Below $4^{\circ}C$,the hydrogen bonding structure causes the water to expand,leading to a decrease in density. Therefore,water has maximum density at $4^{\circ}C$.

(B) The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the external atmospheric pressure.
As we go deep down into a mine,the atmospheric pressure increases due to the weight of the air column above.
Since the boiling point of water increases with an increase in external pressure,the water inside the mine will boil at a temperature greater than $100^{\circ}C$.
Therefore,the correct option is $B$.

(C) At the boiling point,the saturation vapour pressure of a liquid becomes equal to the external atmospheric pressure.
For water,the boiling point at standard atmospheric pressure is $100^{\circ}C$.
Standard atmospheric pressure is defined as $760\, mm$ of $Hg$.
Therefore,the saturation vapour pressure of water at $100^{\circ}C$ is $760\, mm$ of $Hg$.

(A) The calorie is a unit of energy defined as the amount of heat energy required to raise the temperature of $1 \, g$ of water by $1^{\circ} C$.
Specifically,the $15^{\circ} C$ calorie is defined as the heat required to raise the temperature of $1 \, g$ of water from $14.5^{\circ} C$ to $15.5^{\circ} C$ at a standard atmospheric pressure of $760 \, mm$ of $Hg$.

(B) The boiling point of a liquid is defined as the temperature at which its vapor pressure becomes equal to the external atmospheric pressure.
When the external pressure is reduced, the vapor pressure of the liquid reaches the external pressure at a lower temperature.
Therefore, the boiling point of water decreases as the pressure is reduced.
Thus, option $B$ is correct.

(B) The triple point of water is the specific temperature and pressure at which the three phases of water (solid ice,liquid water,and water vapor) coexist in thermodynamic equilibrium.
By international agreement,the triple point of water is defined as $273.16\, K$ (which is equivalent to $0.01^{\circ}C$ or $32.018^{\circ}F$).
Therefore,the correct option is $B$.

(A) The relationship between temperature in Celsius $(C)$ and Fahrenheit $(F)$ is given by the formula: $\frac{C}{100} = \frac{F - 32}{180}$.
Rearranging this to express $F$ in terms of $C$,we get: $F = \frac{9}{5}C + 32$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = F$ and $x = C$:
Here,the slope $m = \frac{9}{5}$ (which is positive) and the $Y-$intercept $c = 32$ (which is positive).
Since the $Y-$intercept is positive,the line intersects the $Y-$axis at a point above the origin. Thus,the graph has a positive intercept on the $Y-$axis.

(A) The relationship between Celsius $(C)$ and Fahrenheit $(F)$ temperatures is given by the formula:
$\frac{C}{5} = \frac{F - 32}{9}$
Rearranging this equation to the form $y = mx + c$ (where $y = C$ and $x = F$):
$C = \frac{5}{9}F - \frac{32 \times 5}{9}$
$C = \frac{5}{9}F - \frac{160}{9}$
$C = \frac{5}{9}F - 17.78$
This equation represents a straight line with a positive slope $(m = \frac{5}{9})$ and a negative $y$-intercept $(c = -17.78)$.
Looking at the provided figure,curve $1$ is the only line that has a positive slope and passes through the negative $C$-axis (negative intercept).
Therefore,curve $1$ represents the correct relationship.

(A) When the temperature of an object is equal to that of the human body,no heat is transferred between the object and the body.
Since the human body is at a constant temperature,if the objects are also at that same temperature,there is no net flow of heat.
Therefore,both the block of wood and the block of metal will feel neither cold nor hot,or they will feel equally neutral,which is the condition described as feeling 'equally cold or hot' (i.e.,no temperature difference is perceived).
Thus,the correct answer is that their temperatures are equal to the temperature of the human body.

(B) When the surface of a lake is exposed to a freezing environment,ice begins to form on the top layer.
Since ice is a poor conductor of heat,it acts as an insulator between the cold air and the liquid water beneath it.
The phase change of water to ice occurs at the freezing point of water.
Therefore,the temperature of the water immediately in contact with the lower surface of the ice layer is maintained at the freezing point of water,which is $0^{\circ}C$.

(C) lake cools from the surface down. Above $4^{\circ}C$,the cooled water at the surface flows to the bottom because of its greater density.
However,when the surface temperature drops below $4^{\circ}C$ (in this case,$2^{\circ}C$),the water near the surface becomes less dense than the warmer water below.
Consequently,the downward flow ceases,and the water at the bottom remains at $4^{\circ}C$ until nearly the entire lake is frozen.

(B) thermocouple is an electrical device consisting of two dissimilar electrical conductors forming an electrical junction.
It operates on the principle of the Seebeck effect,where a temperature-dependent voltage is produced due to the temperature difference between the two junctions of the dissimilar metals.
Therefore,a thermocouple is primarily used to measure the temperature difference between two points or substances.

(A) The relationship between Fahrenheit temperature $(T_F)$ and Celsius temperature $(T_C)$ is given by the equation: $T_F = \frac{9}{5} T_C + 32$.
Comparing this with the equation of a straight line $Y = mX + C$,where $Y = T_F$,$X = T_C$,$m = 9/5$,and $C = 32$.
Since the intercept $C = 32$ is positive,the line has a positive intercept on the $Y$-axis.

(B) The boiling point of water is $100 ^\circ C$ or $212 ^\circ F$.
The initial temperature of the water is $212 ^\circ F$.
The final temperature of the water is $140 ^\circ F$.
The change in temperature on the Fahrenheit scale is $\Delta T_F = 212 ^\circ F - 140 ^\circ F = 72 ^\circ F$.
The relationship between a change in temperature on the Celsius scale $(\Delta T_C)$ and the Fahrenheit scale $(\Delta T_F)$ is given by $\Delta T_C = \frac{5}{9} \Delta T_F$.
Substituting the value of $\Delta T_F$:
$\Delta T_C = \frac{5}{9} \times 72 = 5 \times 8 = 40 ^\circ C$.
Thus,the Centigrade thermometer will show a drop of $40 ^\circ C$.

(C) The relationship between temperature in Celsius $(T_C)$ and Fahrenheit $(T_F)$ is given by $T_F = \frac{9}{5} T_C + 32$.
We know that $T_C = T_K - 273.15$,where $T_K$ is the temperature in Kelvin.
Substituting $T_K = x$ and $T_F = x$,we get:
$x = \frac{9}{5}(x - 273.15) + 32$
$x = 1.8(x - 273.15) + 32$
$x = 1.8x - 491.67 + 32$
$x = 1.8x - 459.67$
$0.8x = 459.67$
$x = \frac{459.67}{0.8} = 574.5875 \approx 574.25$ (using standard approximation $273$ instead of $273.15$ gives $x = 574.25$ exactly).
Thus,$x = 574.25$.

(D) The relationship between a change in temperature on the Celsius scale $(\Delta T_C)$ and the Fahrenheit scale $(\Delta T_F)$ is given by the formula: $\Delta T_F = \frac{9}{5} \Delta T_C$.
Given that the increase in temperature on the Celsius scale is $\Delta T_C = 30^{\circ}C$.
Substituting this value into the formula:
$\Delta T_F = \frac{9}{5} \times 30^{\circ}F$
$\Delta T_F = 9 \times 6^{\circ}F$
$\Delta T_F = 54^{\circ}F$.
Therefore,the increase in temperature on the Fahrenheit scale is $54^{\circ}F$.

(A) The relationship between the Kelvin scale $(T_K)$ and the Celsius scale $(T_C)$ is given by $T_C = T_K - 273.15$. Taking $T_K = 95 \ K$,we get $T_C = 95 - 273 = -178 \ ^\circ C$.
The relationship between the Fahrenheit scale $(T_F)$ and the Celsius scale $(T_C)$ is given by $T_F = \frac{9}{5} T_C + 32$.
Substituting the value of $T_C$:
$T_F = \frac{9}{5}(-178) + 32$
$T_F = -320.4 + 32$
$T_F = -288.4 \ ^\circ F$.
Rounding to the nearest integer,the value is $-288 \ ^\circ F$.

(B) The density of water is maximum at $4 \, ^\circ C$.
To convert this temperature to Fahrenheit,we use the formula: $T_F = \frac{9}{5} T_C + 32$.
Substituting $T_C = 4 \, ^\circ C$ into the formula:
$T_F = \frac{9}{5}(4) + 32$
$T_F = 7.2 + 32$
$T_F = 39.2 \, ^\circ F$.
Thus,the density of water is maximum at $39.2 \, ^\circ F$.

(C) The relationship between the Celsius scale $(T_C)$ and the Fahrenheit scale $(T_F)$ is given by the formula: $T_F = \frac{9}{5} T_C + 32$.
Absolute zero temperature in Celsius is $T_C = -273.15 \ ^\circ C$.
Substituting this value into the formula:
$T_F = \frac{9}{5} (-273.15) + 32$
$T_F = -491.67 + 32$
$T_F = -459.67 \ ^\circ F$.
Rounding to the nearest whole number,we get $T_F \approx -460 \ ^\circ F$.

(A) The Earth's atmosphere acts like a greenhouse,trapping heat through the greenhouse effect. Without the atmosphere,the heat radiated by the Earth would escape directly into space,leading to a significant decrease in the average surface temperature. Therefore,the temperature would be lower than it is currently.

(D) The density of water is anomalous. As water is cooled from room temperature,it contracts and its density increases until it reaches $4 ^oC$.
At $4 ^oC$,water molecules are most closely packed,resulting in maximum density.
Below $4 ^oC$,water begins to expand as it forms a crystalline structure,causing the density to decrease.
Therefore,the density of water is maximum at $4 ^oC$.

(D) The thermoelectric power $P$ is directly proportional to the temperature difference $\theta$ between the junctions, i.e., $P \propto \theta$.
Given, initial temperature $\theta_1 = 80\,\,^{\circ}\text{C}$ and final temperature $\theta_2 = 100\,\,^{\circ}\text{C}$.
The percentage change in thermoelectric power is given by:
$\text{Percentage Change} = \frac{P_2 - P_1}{P_1} \times 100$
Since $P \propto \theta$, we can write:
$\text{Percentage Change} = \frac{\theta_2 - \theta_1}{\theta_1} \times 100$
Substituting the values:
$\text{Percentage Change} = \frac{100 - 80}{80} \times 100$
$\text{Percentage Change} = \frac{20}{80} \times 100$
$\text{Percentage Change} = \frac{1}{4} \times 100 = 25\,\%$.

(B) The relationship between Celsius $(C)$ and Fahrenheit $(F)$ scales is given by the formula: $C = \frac{5}{9}(F - 32)$.
This can be rewritten as: $C = \frac{5}{9}F - \frac{160}{9}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y$ is the temperature in Celsius and $x$ is the temperature in Fahrenheit,the slope $m$ is the coefficient of $F$.
Therefore,the slope $m = \frac{5}{9}$.
Alternatively,using the coordinates of points $A$ and $B$ from the graph:
Point $A$ is $(32, 0)$ and point $B$ is $(212, 100)$.
Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{100 - 0}{212 - 32} = \frac{100}{180} = \frac{5}{9}$.

(C) The relationship between any temperature scale and the Celsius scale is given by the formula: $\frac{X - LFP}{UFP - LFP} = \frac{C - 0^o}{100^o - 0^o}$,where $LFP$ is the Lower Fixed Point and $UFP$ is the Upper Fixed Point.
Given: $LFP = 20^o X$,$UFP = 150^o X$,and $C = 60^o C$.
Substituting the values into the formula:
$\frac{X - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\frac{X - 20}{130} = \frac{60}{100}$
$\frac{X - 20}{130} = 0.6$
$X - 20 = 0.6 \times 130$
$X - 20 = 78$
$X = 78 + 20 = 98^o X$.

(D) The relationship between two linear temperature scales is given by the formula: $\frac{T_W - \text{Freezing point}_W}{\text{Boiling point}_W - \text{Freezing point}_W} = \frac{T_C - \text{Freezing point}_C}{\text{Boiling point}_C - \text{Freezing point}_C}$.
Given for the $W$ scale: Freezing point = $39\,^{\circ}W$,Boiling point = $239\,^{\circ}W$.
Given for the Celsius scale: Freezing point = $0\,^{\circ}C$,Boiling point = $100\,^{\circ}C$.
Substituting the values:
$\frac{T_W - 39}{239 - 39} = \frac{39 - 0}{100 - 0}$
$\frac{T_W - 39}{200} = \frac{39}{100}$
$T_W - 39 = \frac{39}{100} \times 200$
$T_W - 39 = 39 \times 2 = 78$
$T_W = 78 + 39 = 117\,^{\circ}W$.