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: $t = 1.2 \times 100^\circ\text{C} = 120^\circ\text{C}$

Step 1: Identify given values: initial length $l = 20$ cm, change in temperature $\Delta T = 40^\circ C - 20^\circ C = 20^\circ C$, and $\alpha = 1.2 \times 10^{-5} / ^\circ C$.
Step 2: Calculate the increase in length using the formula $\Delta l = l \alpha \Delta T$.
Step 3: Substitute the values to find the expansion: $\Delta l = 20 \times 1.2 \times 10^{-5} \times 20 = 0.48 \times 10^{-2}$ cm.
Step 4: Calculate the actual length of the ruler at $40^\circ C$: $l' = l + \Delta l = 20 + 0.0048 = 20.0048$ cm.

Step 1: Let the length of the brass rod be $l_b$ and the steel rod be $l_s$. Given $l_s - l_b = 5$ cm, so $l_s = l_b + 5$.
Step 2: For the difference in lengths to remain constant at all temperatures, the change in length of both rods must be equal: $\Delta l_s = \Delta l_b$.
Step 3: Apply the linear expansion formula: $l_s \alpha_s \Delta T = l_b \alpha_b \Delta T$.
Step 4: Substitute the given values: $(l_b + 5) \times 12 \times 10^{-6} = l_b \times 18 \times 10^{-6}$.
Step 5: Solve for $l_b$: $12 l_b + 60 = 18 l_b \implies 6 l_b = 60 \implies l_b = 10$ cm.

Step 1: Identify given values: diameter $d = 0.4$ m, so radius $r = 0.2$ m. Initial temperature $T_1 = 273$ K, final temperature $T_2 = 360$ K.
Step 2: Calculate the initial surface area of the ball: $A_0 = 4 \pi r^2 = 4 \times \pi \times (0.2)^2 = 0.5024 m^2$.
Step 3: Calculate the change in temperature: $\Delta T = 360 - 273 = 87$ K.
Step 4: Use the areal expansion formula to find the increase in surface area: $\Delta A = \beta A_0 \Delta T$.
Step 5: Substitute the values: $\Delta A = 0.000034 \times 0.5024 \times 87 = 1.486 \times 10^{-3} m^2$.

Step 1: Identify the given values: initial volume $V = 10000 cm^3$, change in temperature $\Delta T = 40^\circ C - 25^\circ C = 15^\circ C$, and increase in volume $\Delta V = 4 cm^3$.
Step 2: Calculate the coefficient of cubical (volume) expansion using the formula $\gamma = \frac{\Delta V}{V \cdot \Delta T}$.
Step 3: Substitute the values to find $\gamma$: $\gamma = \frac{4}{10000 \times 15} = 26.67 \times 10^{-6} / ^\circ C$.
Step 4: Use the relation between linear expansion coefficient $\alpha$ and volume expansion coefficient $\gamma$, which is $\alpha = \frac{\gamma}{3}$.
Step 5: Calculate the coefficient of linear expansion: $\alpha = \frac{26.67 \times 10^{-6}}{3} = 8.89 \times 10^{-6} / ^\circ C$.

Step 1: Identify the given values: initial volume $V = 10^{-6} m^3$, area of cross-section $A = 2 \times 10^{-7} m^2$, change in temperature $\Delta T = 100^\circ C$, and coefficient of volume expansion $\gamma = 18 \times 10^{-5} / ^\circ C$.
Step 2: Calculate the increase in volume of the mercury using the formula $\Delta V = V \times \gamma \Delta T$.
Step 3: Substitute the values to find $\Delta V$: $\Delta V = 10^{-6} \times 18 \times 10^{-5} \times 100 = 18 \times 10^{-9} m^3$.
Step 4: Relate the increase in volume to the increase in length of the mercury column in the capillary tube using the formula $\Delta V = A \times \Delta l$.
Step 5: Substitute the known values to solve for $\Delta l$: $18 \times 10^{-9} = 2 \times 10^{-7} \times \Delta l$.
Step 6: Calculate the final increase in length: $\Delta l = \frac{18 \times 10^{-9}}{2 \times 10^{-7}} = 9 \times 10^{-2} m = 9 cm$.

Given, coefficient of linear expansion, $\alpha = 12 \times 10^{-6} /^\circ C$
Change in temperature, $\Delta \theta = 40^\circ C - 20^\circ C = 20^\circ C$
Total time in a day, $t = 24 \times 60 \times 60 = 86400$ s
The time lost by the clock in a day is given by the formula:
$\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$
Substituting the given values into the formula:
$\Delta t = \frac{1}{2} \times 12 \times 10^{-6} \times 20 \times 86400$
$\Delta t = 12 \times 10^{-5} \times 86400$
$\Delta t = 10368 \times 10^{-3}$
$\Delta t = 10.368$ s
Rounding to one decimal place, the time lost is $10.4$ s.
Since the temperature increases, the length and time period of the pendulum increase, meaning the clock runs slowly and loses time.
Therefore, the loss in time per day is 10.4 s.

Given, time period of second's pendulum, $T = 2$ s
Calibration temperature, $\theta_1 = 20^\circ C$
Final temperature, $\theta_2 = 30^\circ C$
Change in temperature, $\Delta \theta = 30^\circ C - 20^\circ C = 10^\circ C$
Coefficient of linear expansion of steel, $\alpha = 1.2 \times 10^{-5} /^\circ C$
Total time in one week, $t = 7 \times 24 \times 3600 = 604800$ s
Change in time period is given by the formula:
$\Delta T = \frac{1}{2} T \alpha \Delta \theta$
Substituting the given values:
$\Delta T = \frac{1}{2} \times 2 \times 1.2 \times 10^{-5} \times 10$
$\Delta T = 1.2 \times 10^{-4}$ s
The new time period is:
$T' = T + \Delta T = 2 + 1.2 \times 10^{-4} = 2.00012$ s
Time lost in one week is given by the formula:
$\Delta t = \frac{\Delta T}{T'} \times t$
Substituting the values:
$\Delta t = \frac{1.2 \times 10^{-4}}{2.00012} \times 604800$
$\Delta t = 36.28$ s
Since the time period increases, the clock runs slowly and loses time.
Therefore, the clock loses 36.28 s in one week.