Thermo Dynamics

Step 1: Use the conversion formula between Celsius and Fahrenheit: $\frac{C - 0}{100} = \frac{F - 32}{180}$
Step 2: Substitute F = 60 to find C: $C = \frac{5}{9} (60 - 32) = \frac{5}{9} (28) = 15.55^\circ\text{C}$
Step 3: Convert Celsius to Kelvin using the formula: $K = C + 273.15$
Step 4: Substitute C = 15.55 to find K: $K = 15.55 + 273.15 = 288.7\text{ K}$

Step 1: Calculate the change in temperature in Celsius: $\Delta C = 90^\circ\text{C} - 30^\circ\text{C} = 60^\circ\text{C}$
Step 2: Use the relation for temperature difference on the Fahrenheit scale: $\Delta F = \frac{9}{5} \Delta C$
Step 3: Substitute the value of $\Delta C$: $\Delta F = \frac{9}{5} \times 60 = 108^\circ\text{F}$
Step 4: Use the relation for temperature difference on the Kelvin scale: $\Delta K = \Delta C$
Step 5: Therefore, the change in Kelvin is $\Delta K = 60\text{ K}$

Step 1: Use the general conversion equation for temperature scales: $\frac{T_{known} - LFP_{known}}{UFP_{known} - LFP_{known}} = \frac{T_{unknown} - LFP_{unknown}}{UFP_{unknown} - LFP_{unknown}}$
Step 2: Identify the known values for Celsius: $T_{known} = 39$, $LFP = 0$, $UFP = 100$
Step 3: Identify the unknown values for W scale: $LFP = 39$, $UFP = 239$, $T_{unknown} = t$
Step 4: Substitute the values into the equation: $\frac{39 - 0}{100 - 0} = \frac{t - 39}{239 - 39}$
Step 5: Solve for t: $\frac{39}{100} = \frac{t - 39}{200} \implies t - 39 = 78 \implies t = 117^\circ\text{W}$

Step 1: Apply the standard temperature conversion formula: $\frac{C - 0}{100 - 0} = \frac{t - (-20)}{180 - (-20)}$
Step 2: Substitute the given arbitrary scale reading $t = 5^\circ$: $\frac{C}{100} = \frac{5 + 20}{180 + 20}$
Step 3: Simplify the right side of the equation: $\frac{C}{100} = \frac{25}{200}$
Step 4: Solve for C to find the centigrade reading: $C = \frac{25}{200} \times 100 = 12.5^\circ\text{C}$

Step 1: Identify the lower fixed point (LFP) and upper fixed point (UFP) of the faulty scale: $\theta_0 = 5^\circ$ and $n = 95 - 5 = 90$ divisions.
Step 2: Set up the conversion equation relating the faulty scale reading $\theta$ to the Celsius scale C: $\frac{C - 0}{100} = \frac{\theta - \theta_0}{n}$
Step 3: Substitute the known values $\theta = 59^\circ$, $\theta_0 = 5^\circ$, and $n = 90$: $\frac{C}{100} = \frac{59 - 5}{90}$
Step 4: Simplify the fraction: $\frac{C}{100} = \frac{54}{90}$
Step 5: Solve for C: $C = \frac{54}{90} \times 100 = 60^\circ\text{C}$

Step 1: Identify the given pressures: $p_0 = 66$ cm of Hg at $0^\circ C$, $p_{100} = 88$ cm of Hg at $100^\circ C$, and $p_t = 82.5$ cm of Hg at unknown temperature $t$.
Step 2: Apply the formula for a constant volume gas thermometer: $t = \left(\frac{p_t - p_0}{p_{100} - p_0}\right) \times 100^\circ C$
Step 3: Substitute the values into the formula: $t = \left(\frac{82.5 - 66}{88 - 66}\right) \times 100^\circ C$
Step 4: Simplify the expression to find the temperature: $t = \frac{16.5}{22} \times 100^\circ C = 75^\circ C$

Step 1: Identify the given resistances: $R_0 = 3.70 \Omega$ at $0^\circ C$, $R_{100} = 4.71 \Omega$ at $100^\circ C$, and $R_t = 5.29 \Omega$ at temperature $t$.
Step 2: Apply the temperature formula for a resistance thermometer: $t = 100^\circ C \times \frac{R_t - R_0}{R_{100} - R_0}$
Step 3: Substitute the given values into the equation: $t = 100^\circ C \times \frac{5.29 - 3.70}{4.71 - 3.70}$
Step 4: Calculate the numerator and denominator: $t = 100^\circ C \times \frac{1.59}{1.01}$
Step 5: Solve for the final temperature: $t = 157.4^\circ C$

Step 1: A hygrometer is a device used to measure humidity.
Step 2: A thermometer is a device specifically designed to measure the temperature of a substance.
Step 3: A barometer is used to measure atmospheric pressure.
Step 4: A calorimeter is used to measure the heat of chemical reactions or physical changes.
Step 5: Therefore, the correct device for measuring temperature is a thermometer.

Step 1: The relationship between temperature difference in Celsius ($\Delta T_C$) and Fahrenheit ($\Delta T_F$) is given by the formula $\Delta T_F = \frac{9}{5} \Delta T_C$.
Step 2: Substitute the given temperature difference $\Delta T_C = 25^\circ \mathrm{C}$ into the formula.
Step 3: Calculate the value: $\Delta T_F = \frac{9}{5} \times 25^\circ \mathrm{C}$.
Step 4: Simplify the expression to get $\Delta T_F = 45^\circ \mathrm{F}$.

Step 1: Let the required temperature where Celsius and Fahrenheit scales are equal be $\theta$.
Step 2: Use the standard temperature conversion formula: $\frac{\theta - 0}{100} = \frac{\theta - 32}{180}$.
Step 3: Simplify the equation by reducing fractions: $\frac{\theta}{5} = \frac{\theta - 32}{9}$.
Step 4: Solve for $\theta$ by cross-multiplying: $9\theta = 5\theta - 160$.
Step 5: Rearrange to find $\theta$: $4\theta = -160$, which gives $\theta = -40^\circ$.

Step 1: Use the general temperature conversion formula: $\frac{t - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0}$.
Step 2: Identify the given values: Lower Fixed Point (LFP) $= -20^\circ$, Upper Fixed Point (UFP) $= 130^\circ$, and Celsius temperature $C = 34^\circ \mathrm{C}$.
Step 3: Substitute the values into the formula: $\frac{t - (-20)}{130 - (-20)} = \frac{34 - 0}{100}$.
Step 4: Simplify the equation: $\frac{t + 20}{150} = \frac{34}{100}$.
Step 5: Solve for $t$: $t + 20 = \frac{34 \times 150}{100} = 51$.
Step 6: Therefore, $t = 51 - 20 = 31^\circ$.

Step 1: Use the general temperature conversion formula: $\frac{t - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0}$
Step 2: Identify the given values: Lower Fixed Point (LFP) = -20°, Upper Fixed Point (UFP) = 130°, and Celsius temperature C = 34°C.
Step 3: Substitute the values into the formula: $\frac{t - (-20)}{130 - (-20)} = \frac{34 - 0}{100}$
Step 4: Simplify the equation: $\frac{t + 20}{150} = \frac{34}{100}$
Step 5: Solve for t: $t + 20 = \frac{34 \times 150}{100} = 51$
Step 6: Therefore, $t = 51 - 20 = 31^\circ$

Step 1: Use the standard relationship between Kelvin (K) and Fahrenheit (F) scales: $\frac{F - 32}{9} = \frac{K - 273}{5}$
Step 2: Substitute F = x and K = x as given in the problem: $\frac{x - 32}{9} = \frac{x - 273}{5}$
Step 3: Cross-multiply to solve for x: $5(x - 32) = 9(x - 273)$
Step 4: Expand both sides: $5x - 160 = 9x - 2457$
Step 5: Rearrange the terms: $4x = 2457 - 160 = 2297$
Step 6: Solve for x: $x = \frac{2297}{4} = 574.25$

Step 1: Identify the lower fixed point (LFP) and upper fixed point (UFP) of the faulty scale: $\theta_0 = 5^\circ$ and $\theta_{100} = 95^\circ$.
Step 2: The number of divisions on the faulty scale is $n = 95 - 5 = 90$.
Step 3: Set up the conversion equation relating the faulty scale reading $\theta$ to the Celsius scale C: $\frac{C - 0}{100} = \frac{\theta - \theta_0}{n}$
Step 4: Substitute the known values C = 40°C, $\theta_0 = 5^\circ$, and n = 90: $\frac{40}{100} = \frac{\theta - 5}{90}$
Step 5: Simplify the fraction: $0.4 = \frac{\theta - 5}{90}$
Step 6: Solve for $\theta$: $\theta - 5 = 36 \implies \theta = 41^\circ$

Step 1: Let the reading on the Celsius scale be $T_C$ and on the Fahrenheit scale be $T_F$.
Step 2: According to the given ratio: $\frac{T_C}{T_F} = \frac{2}{5} \implies T_F = \frac{5}{2} T_C$
Step 3: Use the standard temperature conversion formula: $\frac{T_C}{100} = \frac{T_F - 32}{180}$
Step 4: Substitute $T_F = \frac{5}{2} T_C$ into the formula: $\frac{T_C}{100} = \frac{\frac{5}{2} T_C - 32}{180}$
Step 5: Cross-multiply and simplify: $180 T_C = 100 \left( \frac{5}{2} T_C - 32 \right)$
Step 6: Expand and solve for $T_C$: $180 T_C = 250 T_C - 3200 \implies 70 T_C = 3200 \implies T_C = \frac{3200}{70} \approx 45.71^\circ\text{C}$

Step 1: Identify the given readings: $V_0 = 50$ cm at $0^\circ\text{C}$, $V_{100} = 75$ cm at $100^\circ\text{C}$, and $V_t = 80$ cm at unknown temperature t.
Step 2: Apply the formula for a constant pressure gas thermometer: $t = \left( \frac{V_t - V_0}{V_{100} - V_0} \right) \times 100^\circ\text{C}$
Step 3: Substitute the values into the formula: $t = \left( \frac{80 - 50}{75 - 50} \right) \times 100^\circ\text{C}$
Step 4: Simplify the expression: $t = \left( \frac{30}{25} \right) \times 100^\circ\text{C}$
Step 5: Calculate the final temperature: $t = 1.2 \times 100^\circ\text{C} = 120^\circ\text{C}$