Temperature and Temperature scales Questions in English

(D) The relationship between Celsius $(^{\circ}C)$ and Fahrenheit $(^{\circ}F)$ scales is given by:
$\frac{F-32}{180} = \frac{C}{100}$
Let the temperature be $Q$ such that $F = C = Q$.
Substituting $Q$ into the equation:
$\frac{Q-32}{180} = \frac{Q}{100}$
Simplify the fraction $\frac{180}{100}$ to $\frac{9}{5}$:
$Q - 32 = \frac{9}{5}Q$
Rearrange the terms to solve for $Q$:
$Q - \frac{9}{5}Q = 32$
$-\frac{4}{5}Q = 32$
Multiply both sides by $-\frac{5}{4}$:
$Q = 32 \times (-\frac{5}{4})$
$Q = -40$
Thus,the temperature is $-40^{\circ}C$ or $-40^{\circ}F$.

(A) The principle of thermometry states that the ratio of the difference between a temperature and the Lower Fixed Point $(LFP)$ to the difference between the Upper Fixed Point $(UFP)$ and the $LFP$ is constant for all scales.
Let $C$ be the temperature on the Celsius scale and $L$ be the temperature on the new scale.
For the Celsius scale: $LFP = 0^{\circ} C$,$UFP = 100^{\circ} C$.
For the new liquid scale: $LFP = -50^{\circ} C$ (which corresponds to $0^{\circ} L$),$UFP = 150^{\circ} C$ (which corresponds to $100^{\circ} L$).
The conversion formula is: $\frac{L - 0}{100 - 0} = \frac{C - (-50)}{150 - (-50)}$.
$\frac{L}{100} = \frac{C + 50}{200}$.
$L = \frac{C + 50}{2} \Rightarrow 2L = C + 50 \Rightarrow C = 2L - 50$.
For the melting point of water $(C = 0^{\circ} C)$:
$0 = 2L - 50 \Rightarrow 2L = 50 \Rightarrow L = 25^{\circ} L$.
For the boiling point of water $(C = 100^{\circ} C)$:
$100 = 2L - 50 \Rightarrow 2L = 150 \Rightarrow L = 75^{\circ} L$.
Thus,the melting and boiling points of water on this scale are $25^{\circ} L$ and $75^{\circ} L$.

(D) The correct answer is $D$.
$A$ liquid used in a thermometer must be easily visible,expand uniformly and significantly,and remain in the liquid state at room temperature.
Mercury is used because it has a high coefficient of thermal expansion,it is shiny (making it easy to read),and it is a liquid at room temperature.
High density is a physical property of mercury,but it is not a requirement or a reason for its use in clinical thermometers.

(A) Given the relation between the new scale $t_x$ and absolute temperature $T$ is $t_x = 3T + 100$.
The change in temperature $\Delta t_x$ is related to the change in absolute temperature $\Delta T$ by:
$\Delta t_x = t_{x2} - t_{x1} = (3T_2 + 100) - (3T_1 + 100) = 3(T_2 - T_1) = 3 \Delta T$.
Specific heat $c$ is defined as the amount of heat $Q$ required to raise the temperature of unit mass $m$ by unit temperature change $\Delta \theta$:
$c = \frac{Q}{m \Delta \theta}$.
Given $c_x = 1400 \, J \, kg^{-1} X^{-1}$,we have $c_x = \frac{Q}{m \Delta t_x} = 1400$.
We want to find $c_{SI} = \frac{Q}{m \Delta T}$.
Substituting $\Delta t_x = 3 \Delta T$ into the expression for $c_x$:
$1400 = \frac{Q}{m (3 \Delta T)} = \frac{1}{3} \left( \frac{Q}{m \Delta T} \right) = \frac{1}{3} c_{SI}$.
Therefore,$c_{SI} = 3 \times 1400 = 4200 \, J \, kg^{-1} K^{-1}$.

(A) For a constant volume gas thermometer,the temperature $T$ is related to the pressure $P$ by the linear equation: $T = aP + b$.
Given:
At $T_0 = 0^{\circ} C$,$P_0 = 50 \, cm$ of Hg.
At $T_1 = 100^{\circ} C$,$P_1 = 90 \, cm$ of Hg.
The slope of the temperature-pressure graph is $m = \frac{T_1 - T_0}{P_1 - P_0} = \frac{100 - 0}{90 - 50} = \frac{100}{40} = 2.5 \, ^{\circ} C/cm$.
The equation for temperature is $T - T_0 = m(P - P_0)$.
Substituting the values for $P = 60 \, cm$:
$T - 0 = 2.5 \times (60 - 50)$
$T = 2.5 \times 10$
$T = 25^{\circ} C$.

(D) The relationship between the temperature on the centigrade scale $(C)$ and the Fahrenheit scale $(F)$ is given by the formula: $F = \frac{9}{5}C + 32$.
Let the initial temperature be $C_1$ and the final temperature be $C_2$. The increase in temperature on the centigrade scale is $\Delta C = C_2 - C_1 = 30^{\circ}C$.
The corresponding temperatures on the Fahrenheit scale are $F_1 = \frac{9}{5}C_1 + 32$ and $F_2 = \frac{9}{5}C_2 + 32$.
The increase in temperature on the Fahrenheit scale is $\Delta F = F_2 - F_1$.
Substituting the expressions for $F_1$ and $F_2$:
$\Delta F = (\frac{9}{5}C_2 + 32) - (\frac{9}{5}C_1 + 32) = \frac{9}{5}(C_2 - C_1) = \frac{9}{5}(\Delta C)$.
Given $\Delta C = 30$,we have:
$\Delta F = \frac{9}{5} \times 30 = 9 \times 6 = 54^{\circ}F$.
Therefore,the increase in temperature on the Fahrenheit scale is $54^{\circ}F$.

(A) The correct answer is $A$.
Temperature is defined as the physical quantity that measures the degree of hotness or coldness of a body. It determines the direction of heat flow when two bodies are in thermal contact.

(C) For a constant volume gas thermometer,the temperature $T$ is related to the pressure $P$ by the linear relation:
$T = \frac{P - P_0}{P_{100} - P_0} \times 100^{\circ} C$
Given:
$P_0 = 27.50 \, cm$ of $Hg$ at $0^{\circ} C$
$P_{100} = 37.50 \, cm$ of $Hg$ at $100^{\circ} C$
$P = 32.45 \, cm$ of $Hg$ at unknown temperature $T$
Substituting the values:
$T = \frac{32.45 - 27.50}{37.50 - 27.50} \times 100$
$T = \frac{4.95}{10.00} \times 100$
$T = 0.495 \times 100 = 49.5^{\circ} C$
Thus,the unknown temperature is $49.5^{\circ} C$.

(B) Let the temperature at the $60 \, cm$ mark be $T$.
Since the scale is uniform,the temperature varies linearly with the length.
The relationship between temperature $T$ and length $L$ is given by the linear interpolation formula:
$\frac{T - T_0}{T_{100} - T_0} = \frac{L - L_0}{L_{100} - L_0}$
Here,$T_0 = 20^{\circ} C$ at $L_0 = 0 \, cm$ and $T_{100} = 100^{\circ} C$ at $L_{100} = 100 \, cm$.
Substituting the values for $L = 60 \, cm$:
$\frac{T - 20}{100 - 20} = \frac{60 - 0}{100 - 0}$
$\frac{T - 20}{80} = \frac{60}{100}$
$\frac{T - 20}{80} = 0.6$
$T - 20 = 0.6 \times 80$
$T - 20 = 48$
$T = 48 + 20 = 68^{\circ} C$
Thus,the temperature at the $60 \, cm$ mark is $68^{\circ} C$.

(B) The general formula for converting between two temperature scales is given by:
$\frac{\text{Reading on scale} - \text{Lower fixed point}}{\text{Upper fixed point} - \text{Lower fixed point}} = \text{Constant}$
From the graph,for scale $P$,the lower fixed point is $30$ and the upper fixed point is $180$ (since $180 - 30 = 150$ divisions).
For scale $Q$,the lower fixed point is $0$ and the upper fixed point is $100$ (since $100 - 0 = 100$ divisions).
Applying the formula:
$\frac{t_P - 30}{180 - 30} = \frac{t_Q - 0}{100 - 0}$
Simplifying the expression:
$\frac{t_P - 30}{150} = \frac{t_Q}{100}$

(C) Let $T_f$ be the reading on the faulty thermometer and $T_c$ be the reading on the Celsius scale.
The formula for conversion is $\frac{T_f - LFP}{UFP - LFP} = \frac{T_c - 0}{100 - 0}$,where $LFP$ is the Lower Fixed Point and $UFP$ is the Upper Fixed Point.
Given $LFP = 5^{\circ}C$ and $UFP = 95^{\circ}C$,we have $\frac{41 - 5}{95 - 5} = \frac{T_c}{100}$.
$\frac{36}{90} = \frac{T_c}{100}$.
$T_c = \frac{36}{90} \times 100 = 40^{\circ}C$.
To convert to the absolute scale (Kelvin),$T_K = T_c + 273.15 \approx 40 + 273 = 313 \, K$.

(D) The relationship between any linear temperature scale and the Fahrenheit scale is given by the formula:
$\frac{X - X_{\text{freezing}}}{X_{\text{boiling}} - X_{\text{freezing}}} = \frac{F - 32}{212 - 32}$
Given:
$X_{\text{boiling}} = 65^{\circ} X$
$X_{\text{freezing}} = -15^{\circ} X$
$X = -95^{\circ} X$
Substituting the values into the formula:
$\frac{-95 - (-15)}{65 - (-15)} = \frac{F - 32}{180}$
$\frac{-95 + 15}{65 + 15} = \frac{F - 32}{180}$
$\frac{-80}{80} = \frac{F - 32}{180}$
$-1 = \frac{F - 32}{180}$
$-180 = F - 32$
$F = -180 + 32 = -148^{\circ} F$

(C) 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 $\Delta T_C = 40^{\circ} C$.
Substituting this value into the formula:
$\Delta T_F = \frac{9}{5} \times 40^{\circ} F$
$\Delta T_F = 9 \times 8^{\circ} F$
$\Delta T_F = 72^{\circ} F$.
Therefore,the increase in temperature on the Fahrenheit scale is $72^{\circ} F$.
Hence,the correct option is $C$.

(B) The relationship between Celsius $(C)$ and Fahrenheit $(F)$ scales 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 $x = F$):
$C = \frac{5}{9}F - \frac{160}{9}$
This is a linear equation of the form $y = mx + c$,where the slope $m = \frac{5}{9}$ is positive and the $y$-intercept $c = -\frac{160}{9}$ is negative.
$A$ graph with a positive slope and a negative $y$-intercept corresponds to a line that passes through the fourth quadrant and has a positive inclination.
Comparing this with the given options,Figure $B$ represents a line with a positive slope and a negative $y$-intercept.

(C) The general formula for temperature conversion between any two scales is: $\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{Y - \text{LFP}_Y}{\text{UFP}_Y - \text{LFP}_Y}$.
For the faulty thermometer $(X)$: $\text{LFP}_X = 5^{\circ} C$,$\text{UFP}_X = 99^{\circ} C$,and reading $X = 52^{\circ} C$.
For the Fahrenheit scale $(Y)$: $\text{LFP}_Y = 32^{\circ} F$,$\text{UFP}_Y = 212^{\circ} F$.
Substituting the values: $\frac{52 - 5}{99 - 5} = \frac{T_F - 32}{212 - 32}$.
$\frac{47}{94} = \frac{T_F - 32}{180}$.
$0.5 = \frac{T_F - 32}{180}$.
$T_F - 32 = 90$.
$T_F = 122^{\circ} F$.

(C) The relationship between the Kelvin scale $(K)$ and the Fahrenheit scale $(F)$ is given by the formula: $\frac{K - 273.15}{5} = \frac{F - 32}{9}$.
Given that the temperature is $x$ on both scales,we substitute $K = x$ and $F = x$ into the equation:
$\frac{x - 273.15}{5} = \frac{x - 32}{9}$.
Cross-multiplying gives: $9(x - 273.15) = 5(x - 32)$.
$9x - 2458.35 = 5x - 160$.
$4x = 2458.35 - 160$.
$4x = 2298.35$.
$x = \frac{2298.35}{4} \approx 574.58$.
Rounding to the nearest whole number,we get $x \approx 574$.

(B) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta T$.
Given: $V_0 = 10^{-6} \,m^3$, $\gamma = 18 \times 10^{-5} /{ }^{\circ} C$, and $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.
Substituting the values: $\Delta V = (10^{-6} \,m^3) \times (18 \times 10^{-5} /{ }^{\circ} C) \times (100^{\circ} C) = 18 \times 10^{-9} \,m^3$.
The cross-sectional area of the stem $A = 0.002 \,cm^2 = 0.002 \times 10^{-4} \,m^2 = 2 \times 10^{-7} \,m^2$.
The change in volume is also equal to the area times the length of the mercury column: $\Delta V = A \times L$.
Therefore, $L = \frac{\Delta V}{A} = \frac{18 \times 10^{-9} \,m^3}{2 \times 10^{-7} \,m^2} = 9 \times 10^{-2} \,m = 9 \,cm$.

(B) The relationship between the temperature in Celsius $(C)$ and Fahrenheit $(F)$ scales is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
Given the Fahrenheit temperature $F = 140^{\circ} F$.
Substituting the value of $F$ into 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 temperature registered by the centigrade thermometer is $60^{\circ} C$.

(C) In the Celsius scale,the freezing point of water is $0^{\circ} C$ and the boiling point is $100^{\circ} C$. The range is $100 - 0 = 100$.
In the given imaginary '$W$' scale,the freezing point is $39^{\circ} W$ and the boiling point is $239^{\circ} W$. The range is $239 - 39 = 200$.
Using the linear conversion formula between two temperature scales:
$\frac{C - C_{freezing}}{C_{boiling} - C_{freezing}} = \frac{W - W_{freezing}}{W_{boiling} - W_{freezing}}$
Substituting the values:
$\frac{C - 0}{100 - 0} = \frac{W - 39}{239 - 39}$
$\frac{C}{100} = \frac{W - 39}{200}$
For $C = 39^{\circ} C$:
$\frac{39}{100} = \frac{W - 39}{200}$
$39 \times 2 = W - 39$
$78 = W - 39$
$W = 78 + 39 = 117^{\circ} W$.

(B) The relation between temperature in Celsius $(C)$ and Kelvin $(K)$ is given by:
$T(K) = T(^{\circ}C) + 273.15$
Given $T(^{\circ}C) = -197^{\circ}C$.
Substituting the value:
$T(K) = -197 + 273 = 76 K$
Therefore,the temperature is $76 K$.

(C) Given:
Resistance at ice point $(R_0)$ = $5 \Omega$ at $\theta_0 = 0^{\circ} C$
Resistance at steam point $(R_{100})$ = $5.23 \Omega$ at $\theta_{100} = 100^{\circ} C$
Resistance in hot bath $(R_{\theta})$ = $5.795 \Omega$
The formula for temperature in a platinum resistance thermometer is:
$\theta = \frac{R_{\theta} - R_0}{R_{100} - R_0} \times 100^{\circ} C$
Substituting the values:
$\theta = \frac{5.795 - 5}{5.23 - 5} \times 100^{\circ} C$
$\theta = \frac{0.795}{0.23} \times 100^{\circ} C$
$\theta = 3.4565 \times 100^{\circ} C$
$\theta = 345.65^{\circ} C$
Therefore,the temperature of the bath is $345.65^{\circ} C$.

(A) Anomalous expansion of water is a unique property where water contracts instead of expanding when heated between $ 0^{\circ} C $ and $ 4^{\circ} C $.
Since density is defined as mass per unit volume $( \rho = m/V )$, as the volume decreases while the mass remains constant, the density increases.
At $ 4^{\circ} C $, the volume of a given mass of water is minimum.
Therefore, the density of water is maximum at $ 4^{\circ} C $.

(B) In thermodynamics, the triple point of water is defined as the specific temperature and pressure at which the three phases of water (liquid, solid, and gaseous) coexist in thermodynamic equilibrium.
By international agreement, the triple point of pure water is exactly $273.16 \,K$ at a pressure of $611.2 \,Pa$.

(C) The relationship between temperature in Fahrenheit $(F)$ and Celsius $(C)$ is given by the formula: $\frac{F - 32}{9} = \frac{C}{5}$.
Given that the ratio of the temperature values is $F:C = 13:5$,we can write $F = 13x$ and $C = 5x$ for some constant $x$.
Substituting these into the formula: $\frac{13x - 32}{9} = \frac{5x}{5}$.
This simplifies to: $\frac{13x - 32}{9} = x$.
Multiplying both sides by $9$: $13x - 32 = 9x$.
Rearranging the terms: $13x - 9x = 32$,which gives $4x = 32$,so $x = 8$.
Therefore,the temperature in Fahrenheit is $F = 13 \times 8 = 104^{\circ} F$ and the temperature in Celsius is $C = 5 \times 8 = 40^{\circ} C$.
Since $40^{\circ} C$ is one of the options,the correct answer is $C$.

(A) Water exhibits anomalous expansion. Its density is maximum at $4^{\circ} C$.
When the surface water cools to $0^{\circ} C$ and begins to freeze,the denser water at $4^{\circ} C$ settles at the bottom of the lake.
Therefore,the temperature at the bottom of the lake remains $4^{\circ} C$.

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

(D) The relation between the Celsius scale temperature $(C)$ and the Fahrenheit scale temperature $(F)$ is given by:
$\frac{C}{5} = \frac{F-32}{9}$
Rearranging this equation to the form $y = mx + c$ (where $y = C$ and $x = F$):
$C = \frac{5}{9}F - \frac{160}{9}$
Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $m = \tan \theta = \frac{5}{9}$.
Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{9}$,we can consider a right-angled triangle with opposite side $5$ and adjacent side $9$.
The hypotenuse is $\sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}$.
Therefore,the cosine of the angle $\theta$ is $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{9}{\sqrt{106}}$.

(D) Let the resistance be a linear function of temperature: $R_T = R_0 + \alpha T$.
Given: $R_1 = 100 \ \Omega$ at $T_1 = 273 \ K$ and $R_2 = 300 \ \Omega$ at $T_2 = 873 \ K$.
The slope of the resistance-temperature graph is $m = \frac{R_2 - R_1}{T_2 - T_1} = \frac{300 - 100}{873 - 273} = \frac{200}{600} = \frac{1}{3} \ \Omega/K$.
We want to find the temperature $T$ where $R = 200 \ \Omega$.
Using the equation of a line: $R - R_1 = m(T - T_1)$.
$200 - 100 = \frac{1}{3}(T - 273)$.
$100 = \frac{1}{3}(T - 273)$.
$300 = T - 273$.
$T = 300 + 273 = 573 \ K$.

(A) thermocouple consists of two junctions of dissimilar metals.
Because the junctions are very small,their thermal capacity (mass $\times$ specific heat) is extremely low.
Thermal capacity determines how much heat is required to change the temperature of an object.
Since the thermal capacity is very small,the junction can quickly gain or lose heat to reach thermal equilibrium with the surrounding environment.
This allows the thermocouple to respond rapidly to changes in temperature.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(C) Let the reading on the Celsius scale be $C$ and the reading on the Fahrenheit scale be $F$.
According to the problem,$F = C + 0.90C = 1.9C$.
The relationship between the Celsius and Fahrenheit scales is given by the formula: $F = \frac{9}{5}C + 32$.
Substituting $F = 1.9C$ into the formula,we get: $1.9C = 1.8C + 32$.
Subtracting $1.8C$ from both sides,we get: $0.1C = 32$.
Thus,$C = 320^{\circ} C$.
Now,find $F$ using $F = 1.9C$: $F = 1.9 \times 320 = 608^{\circ} F$.
Therefore,the temperature is $608^{\circ} F$.

(B) First,convert the temperature measured by the correct Fahrenheit thermometer into Celsius scale.
The relation between Celsius $(C)$ and Fahrenheit $(F)$ is given by: $C = \frac{5}{9}(F - 32)$.
Substituting $F = 122^{\circ} F$:
$C = \frac{5}{9}(122 - 32) = \frac{5}{9}(90) = 50^{\circ} C$.
This is the true temperature of the body.
The faulty thermometer reads $49^{\circ} C$.
The correction is defined as: $\text{True Value} - \text{Measured Value}$.
Correction $= 50^{\circ} C - 49^{\circ} C = +1^{\circ} C$.
Therefore,the correction to be applied is $+1^{\circ} C$.

(D) The relationship between the Fahrenheit scale $(F)$ and the Celsius scale $(C)$ is given by the formula: $\frac{F-32}{9} = \frac{C}{5}$.
Given that the reading in Fahrenheit is twice the reading in Celsius,we have $F = 2C$,which implies $C = \frac{F}{2}$.
Substituting $C = \frac{F}{2}$ into the temperature conversion formula:
$\frac{F-32}{9} = \frac{F/2}{5}$
$\frac{F-32}{9} = \frac{F}{10}$
Cross-multiplying gives: $10(F - 32) = 9F$.
$10F - 320 = 9F$.
$10F - 9F = 320$.
$F = 320^{\circ} F$.

(C) The relationship between the Fahrenheit $(F)$ and Kelvin $(K)$ scales is given by the formula:
$\frac{F - 32}{9} = \frac{K - 273.15}{5}$
Let the temperature where both scales have the same numerical value be $X$.
Substituting $F = X$ and $K = X$ into the equation:
$\frac{X - 32}{9} = \frac{X - 273.15}{5}$
Cross-multiplying gives:
$5(X - 32) = 9(X - 273.15)$
$5X - 160 = 9X - 2458.35$
Rearranging the terms to solve for $X$:
$9X - 5X = 2458.35 - 160$
$4X = 2298.35$
$X = 574.5875 \approx 574.6$
Thus,the temperature is $574.6^{\circ} F$.

(C) The relationship between Fahrenheit $(F)$ and Celsius $(C)$ temperature scales is given by the formula: $F = \frac{9}{5} C + 32$.
According to the problem,the temperature on the Fahrenheit scale is twice the temperature on the Celsius scale,so $F = 2C$,which implies $C = \frac{F}{2}$.
Substituting this into the formula: $F = \frac{9}{5} \left( \frac{F}{2} \right) + 32$.
$F = \frac{9F}{10} + 32$.
Multiplying the entire equation by $10$: $10F = 9F + 320$.
$10F - 9F = 320$.
Therefore,$F = 320^{\circ} F$.

(B) The relationship between two temperature scales is given by the formula:
$\frac{\text{Reading on scale} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{constant}$
For the Celsius scale,the Lower Fixed Point $(LFP)$ is $0^{\circ} C$ and the Upper Fixed Point $(UFP)$ is $100^{\circ} C$.
For the new scale $X$,the $LFP$ is $20^{\circ} X$ and the $UFP$ is $110^{\circ} X$.
Let the temperature on the new scale be $z^{\circ} X$ corresponding to $40^{\circ} C$.
$\frac{z - 20}{110 - 20} = \frac{40 - 0}{100 - 0}$
$\frac{z - 20}{90} = \frac{40}{100}$
$\frac{z - 20}{90} = 0.4$
$z - 20 = 0.4 \times 90$
$z - 20 = 36$
$z = 36 + 20 = 56^{\circ} X$
Therefore,$40^{\circ} C$ corresponds to $56^{\circ} X$.

(C) The relationship between any temperature scale and the Kelvin scale is given by the formula: $\frac{Y - Y_{ice}}{Y_{steam} - Y_{ice}} = \frac{K - K_{ice}}{K_{steam} - K_{ice}}$.
Given for scale $Y$: $Y_{ice} = -160^{\circ} Y$ and $Y_{steam} = -50^{\circ} Y$.
For the Kelvin scale: $K_{ice} = 273 \ K$ and $K_{steam} = 373 \ K$.
Substituting the values for $K = 340 \ K$:
$\frac{Y - (-160)}{-50 - (-160)} = \frac{340 - 273}{373 - 273}$
$\frac{Y + 160}{110} = \frac{67}{100}$
$Y + 160 = 0.67 \times 110$
$Y + 160 = 73.7$
$Y = 73.7 - 160 = -86.3^{\circ} Y$.

(D) The ice point on the Celsius scale is $0^{\circ} C$ and the steam point is $100^{\circ} C$.
In the new scale,the ice point is $25 X$ and the steam point is $305 X$.
The temperature interval of $100^{\circ} C$ corresponds to $(305 - 25) X = 280 X$.
Therefore,a change of $1^{\circ} C$ is equivalent to a change of $\frac{280}{100} X = 2.8 X$.
This implies $1^{\circ} C = 2.8 X$,or $1 X = \frac{1}{2.8}^{\circ} C$.
The specific heat capacity of water is $c = 4200 \ J kg^{-1} (^{\circ} C)^{-1}$.
Substituting the relation $1^{\circ} C = 2.8 X$ into the unit:
$c = 4200 \ J kg^{-1} (2.8 X)^{-1} = \frac{4200}{2.8} J kg^{-1} X^{-1} = 1500 \ J kg^{-1} X^{-1}$.
Thus,$c = 1.5 \times 10^{3} J kg^{-1} X^{-1}$.

(D) The relationship between any temperature scale and the Celsius scale is given by the formula: $\frac{X - \text{Lower Fixed Point}}{\text{Upper Fixed Point} - \text{Lower Fixed Point}} = \frac{C}{100}$.
Here, the lower fixed point is $10^{\circ}$ and the upper fixed point is $130^{\circ}$.
Substituting the given values: $\frac{X - 10}{130 - 10} = \frac{40}{100}$.
$\frac{X - 10}{120} = \frac{40}{100}$.
$X - 10 = \frac{40}{100} \times 120$.
$X - 10 = 0.4 \times 120 = 48$.
$X = 48 + 10 = 58^{\circ}$.

(C) Given the relationship between the two temperature scales is $\frac{A-42}{110} = \frac{B-72}{220}$.
To find the temperature where both scales have the same reading,we set $A = B = T$.
Substituting $T$ into the equation: $\frac{T-42}{110} = \frac{T-72}{220}$.
Simplifying the equation by dividing the denominators by $110$: $\frac{T-42}{1} = \frac{T-72}{2}$.
Cross-multiplying gives: $2(T-42) = T-72$.
Expanding the terms: $2T - 84 = T - 72$.
Rearranging to solve for $T$: $2T - T = 84 - 72$.
Thus,$T = 12$.
Therefore,the two scales have the same reading at $12^{\circ}$.

(C) At high altitudes,the atmospheric pressure is significantly lower than at sea level.
Since the boiling point of a liquid depends on the external pressure,a decrease in atmospheric pressure leads to a decrease in the boiling point of water.
Consequently,water boils at a temperature lower than $100 \, ^\circ\text{C}$ at high altitudes.
Because the water boils at a lower temperature,it does not provide enough heat to cook the rice properly,making the cooking process difficult.