Step 1: Given Variables & Physical Constants

• Base area of the boiler ($A$): $0.15 \, \text{m}^2$
• Thickness of the boiler ($d$): $1.0 \, \text{cm} = 1.0 \times 10^{-2} \, \text{m}$
• Rate of boiling water ($\frac{dm}{dt}$): $6.0 \, \text{kg/min} = \frac{6.0}{60} \, \text{kg/s} = 0.1 \, \text{kg/s}$
• Thermal conductivity of brass ($K$): $109 \, \text{J s}^{-1} \text{m}^{-1} \text{K}^{-1}$
• Latent heat of vaporisation of water ($L$): $2256 \times 10^3 \, \text{J/kg}$
• Temperature of boiling water ($T_2$): $100^\circ\text{C} = 373 \, \text{K}$ [Since water boils at $100^\circ\text{C}$ at atmospheric pressure]

Step 2: Determining the Rate of Heat Flow

The rate of heat transfer ($Q/t$) required to vaporize water at a steady rate is given by the product of the mass flow rate and the latent heat of vaporization [Per the principle of calorimetry]:

$\frac{Q}{t} = \left( \frac{dm}{dt} \right) \times L$

$\frac{Q}{t} = 0.1 \, \text{kg/s} \times 2256 \times 10^3 \, \text{J/kg} = 225,600 \, \text{J/s (or Watts)}$

Step 3: Applying Fourier’s Law of Heat Conduction

The heat transfer rate through a slab of material is governed by the conduction equation [Fourier's Law of Heat Conduction]:

$\frac{Q}{t} = \frac{K \cdot A \cdot (T_1 - T_2)}{d}$

Where $T_1$ is the temperature of the flame side and $T_2$ is the temperature of the water side.

Step 4: Algebraic Derivation for Flame Temperature ($T_1$)

Rearranging the equation to solve for $T_1$:

$T_1 - T_2 = \frac{(\frac{Q}{t}) \cdot d}{K \cdot A}$

$T_1 = T_2 + \frac{(\frac{Q}{t}) \cdot d}{K \cdot A}$

Substituting the known values:

$T_1 = 373 + \frac{225,600 \times 1.0 \times 10^{-2}}{109 \times 0.15}$

$T_1 = 373 + \frac{2256}{16.35}$

$T_1 \approx 373 + 137.98$

$T_1 \approx 510.98 \, \text{K}$

Step 5: Conversion to Celsius

Converting back to Celsius [Scale conversion $T(^\circ\text{C}) = T(\text{K}) - 273.15$]:

$T_1 (^\circ\text{C}) \approx 510.98 - 273 = 237.98^\circ\text{C}$

Rounding to significant figures, we obtain $238^\circ\text{C}$.

Final Solution: The estimated temperature of the flame in contact with the boiler is approximately 511 K or 238°C.

1. Analysis of the Triple Point of Water

The triple point of water is defined as the unique state where water coexists in three phases—solid (ice), liquid (water), and gas (water vapor)—in thermodynamic equilibrium. Its value is internationally fixed at $273.16 \, K$.

Justification for Selection:

• Uniqueness: Per the Gibbs Phase Rule ($F = C - P + 2$), the triple point occurs at a specific, invariant temperature and pressure. For water ($C=1$), at $P=3$ phases, the degree of freedom ($F$) is $0$. This ensures the point is unique and reproducible across all laboratories globally.
• Precision: Unlike phase transitions that occur over a temperature range due to impurity sensitivity, the triple point is a singular point on the $P-T$ phase diagram.

2. Critical Analysis of Original Fixed Points

The original Celsius scale utilized the melting point of ice ($0^\circ C$) and the boiling point of water ($100^\circ C$) as standard fixed points. This methodology possesses inherent physical drawbacks:

3. Logical Conclusion

The adoption of the triple point of water as a single fixed point—in conjunction with the value of the Boltzmann constant—obviates the need for a second reference point. It eliminates the experimental uncertainties introduced by pressure variations and chemical impurities that rendered the two-point system (freezing and boiling) unreliable for modern high-precision metrology.

Final Solution: The triple point of water is chosen because it represents a unique, invariant state of matter defined by fixed pressure and temperature, independent of experimental fluctuations. Conversely, the melting and boiling points are unsuitable as primary standards because they are highly dependent on external pressure and the purity of the substance, leading to lack of international reproducibility.

Given Variables & Initial Setup

To determine the fractional change in density of glycerine, we define the following parameters provided in the problem statement:

• Coefficient of volume expansion of glycerine, $\gamma = 49 \times 10^{-5} \, K^{-1}$
• Change in temperature, $\Delta T = 30 \, °C = 30 \, K$ [As the magnitude of a temperature interval is identical in Celsius and Kelvin scales].
• Initial density of glycerine: $\rho_0$
• Final density of glycerine: $\rho$

Step 1: Establishing the Relationship between Density and Temperature

The volume $V$ of a substance at a temperature $T + \Delta T$ is given by the equation of thermal expansion [Per the definition of the coefficient of volume expansion]:

$V = V_0(1 + \gamma \Delta T)$

Since density $\rho$ is inversely proportional to volume for a fixed mass $m$ ($ \rho = \frac{m}{V} $), the density at the final temperature is:

$\rho = \frac{m}{V} = \frac{m}{V_0(1 + \gamma \Delta T)} = \rho_0(1 + \gamma \Delta T)^{-1}$

Step 2: Applying the Binomial Approximation

Given that the product $\gamma \Delta T$ is very small compared to $1$, we apply the Binomial Theorem $(1 + x)^n \approx 1 + nx$ for $|x| \ll 1$:

$\rho \approx \rho_0(1 - \gamma \Delta T)$

Rearranging this to find the change in density ($\Delta \rho = \rho - \rho_0$):

$\rho - \rho_0 = -\rho_0 \gamma \Delta T$

$\frac{\Delta \rho}{\rho_0} = -\gamma \Delta T$

Step 3: Numerical Calculation

Substitute the known values into the derived expression for fractional change ($\frac{\Delta \rho}{\rho_0}$):

$\text{Fractional change} = -(49 \times 10^{-5} \, K^{-1}) \times (30 \, K)$

$\text{Fractional change} = -(1470 \times 10^{-5})$

$\text{Fractional change} = -1.47 \times 10^{-2}$

Visual Representation of Density Variation

Conclusion on the Negative Sign

The negative sign indicates a decrease in density as temperature increases, which is physically consistent with the expansion of the fluid occupying a larger volume for the same mass [Per the Law of Conservation of Mass].

Final Solution: The fractional change in density is $-1.47 \times 10^{-2}$ (or a decrease of approximately $1.47\%$).

Step 1: Given Variables & Initial Setup

The resistance of the thermometer is defined by the linear relation:

$R = R_0 [1 + \alpha (T - T_0)]$

Where:

• $T_0 = 273.16\text{ K}$ (Triple-point of water)
• $R_0 = 101.6\text{ }\Omega$ (Resistance at $T_0$)
• $T_1 = 600.5\text{ K}$ (Normal melting point of lead)
• $R_1 = 165.5\text{ }\Omega$ (Resistance at $T_1$)
• $R = 123.4\text{ }\Omega$ (Target resistance)

Step 2: Calculating the Temperature Coefficient ($\alpha$)

Using the values at the melting point of lead, we isolate $\alpha$ [Per the linear resistance-temperature model]:

$165.5 = 101.6 [1 + \alpha (600.5 - 273.16)]$

$\frac{165.5}{101.6} = 1 + \alpha (327.34)$

$1.628937 = 1 + \alpha (327.34)$

$0.628937 = \alpha (327.34)$

$\alpha = \frac{0.628937}{327.34} \approx 1.92135 \times 10^{-3}\text{ K}^{-1}$

Step 3: Solving for Unknown Temperature ($T$)

Using the target resistance $R = 123.4\text{ }\Omega$, we rearrange the initial equation to solve for $T$:

$R = R_0 + R_0 \alpha (T - T_0)$

$T - T_0 = \frac{R - R_0}{R_0 \alpha}$

$T = T_0 + \frac{R - R_0}{R_0 \alpha}$

Substitute the known values:

$T = 273.16 + \frac{123.4 - 101.6}{101.6 \times 1.92135 \times 10^{-3}}$

$T = 273.16 + \frac{21.8}{0.195209}$

$T = 273.16 + 111.675$

$T \approx 384.835\text{ K}$

Visual Representation of Resistance-Temperature Gradient

Final Solution: The temperature corresponding to a resistance of 123.4 $\Omega$ is 384.84 K.

Step 1: Identifying Given Variables and Constants

• Power of the drilling machine ($P$): $10\text{ kW} = 10 \times 10^3\text{ W} = 10,000\text{ J/s}$
• Mass of the aluminium block ($m$): $8.0\text{ kg} = 8,000\text{ g}$
• Time duration ($\Delta t$): $2.5\text{ minutes} = 2.5 \times 60\text{ s} = 150\text{ s}$
• Efficiency ($\eta$): $50\% = 0.5$ (fraction of power used for heating)
• Specific heat capacity of aluminium ($c$): $0.91\text{ J g}^{-1}\text{ K}^{-1}$

Step 2: Calculating Total Energy Generated and Utilized

The total energy ($E_{total}$) produced by the machine is given by the product of power and time [By Definition of Work and Energy, $E = P \times \Delta t$]:

$E_{total} = P \times \Delta t = 10,000\text{ J/s} \times 150\text{ s} = 1,500,000\text{ J}$

Since only $50\%$ of the power is effective in heating the aluminium block ($Q_{absorbed}$):

$Q_{absorbed} = \eta \times E_{total}$

$Q_{absorbed} = 0.5 \times 1,500,000\text{ J} = 750,000\text{ J}$

Step 3: Relating Heat to Temperature Change

The heat energy absorbed by a substance is related to the temperature rise ($\Delta T$) by the formula [Per the Principle of Calorimetry]:

$Q = m \cdot c \cdot \Delta T$

Rearranging the equation to solve for the change in temperature ($\Delta T$):

$\Delta T = \frac{Q_{absorbed}}{m \cdot c}$

Step 4: Algebraic Substitution and Calculation

Substitute the known values into the equation:

$\Delta T = \frac{750,000\text{ J}}{8,000\text{ g} \times 0.91\text{ J g}^{-1}\text{ K}^{-1}}$

$\Delta T = \frac{750,000}{7,280}\text{ K}$

$\Delta T \approx 103.02\text{ K}$

Visual Representation of Thermal Energy Transfer

Final Solution: The rise in temperature of the aluminium block is approximately 103.02 K (or 103.02 °C).

Step 1: Given Variables & Parameters

• Standard temperature ($T_0$): $27.0\ ^\circ\text{C}$
• Observed temperature ($T$): $45.0\ ^\circ\text{C}$
• Change in temperature ($\Delta T$): $T - T_0 = 45.0 - 27.0 = 18.0\ \text{K}$ [Since a change in Celsius is equivalent to a change in Kelvin]
• Coefficient of linear expansion of steel ($\alpha$): $1.20 \times 10^{-5}\ \text{K}^{-1}$
• Calibrated length of tape ($L_0$): $1.00\ \text{m} = 100\ \text{cm}$
• Measured length of rod by the tape at $45.0\ ^\circ\text{C}$ ($L_{measured}$): $63.0\ \text{cm}$

Step 2: Calculating the Actual Length of the Rod at 45.0 °C

The steel tape expands at $45.0\ ^\circ\text{C}$. The length of the tape $L'$ corresponding to the $1\ \text{cm}$ marking at $T_0$ is given by the linear expansion formula: $L' = L_0(1 + \alpha \Delta T)$.

Therefore, a marking of $1\ \text{cm}$ on the tape at $45.0\ ^\circ\text{C}$ represents an actual length of:

$L_{actual\_unit} = 1 \times (1 + (1.20 \times 10^{-5} \times 18.0))$

$L_{actual\_unit} = 1 + 2.16 \times 10^{-4} = 1.000216\ \text{cm}$

The actual length of the rod ($L_{actual}$) is the measured length multiplied by this factor:

$L_{actual} = 63.0 \times 1.000216 \approx 63.0136\ \text{cm}$

Step 3: Calculating the Length of the Rod at 27.0 °C

The rod, like the tape, also has a coefficient of linear expansion $\alpha$. Let $L_{27}$ be the length of the rod at $27.0\ ^\circ\text{C}$. The length of the rod at $45.0\ ^\circ\text{C}$ ($L_{45}$) relates to its length at $27.0\ ^\circ\text{C}$ by the expansion formula: $L_{45} = L_{27}(1 + \alpha \Delta T)$.

$63.0136 = L_{27}(1 + (1.20 \times 10^{-5} \times 18.0))$

$63.0136 = L_{27}(1.000216)$

$L_{27} = \frac{63.0136}{1.000216} \approx 63.0000\ \text{cm}$

Visual Representation of Thermal Expansion

Analytical Summary

Final Solution: The actual length of the rod at 45.0 °C is 63.0136 cm, and the length of the same rod at 27.0 °C is 63.0000 cm.

Given Variables & Initial Setup

To determine the mass of ice remaining, we must calculate the total heat energy ($Q$) that flows into the icebox through its walls via thermal conduction over the specified duration.

• Side of the cubical box ($L$): $0.30 \, m$
• Thickness of the walls ($\Delta x$): $0.05 \, m$
• Initial mass of ice ($M_{initial}$): $4.0 \, kg$
• Time duration ($t$): $6 \, hours = 6 \times 3600 \, s = 21600 \, s$
• Outside temperature ($T_2$): $45 \, ^\circ C$
• Inside temperature ($T_1$): $0 \, ^\circ C$ (since ice is present)
• Temperature difference ($\Delta T$): $45 - 0 = 45 \, K$
• Thermal conductivity ($K$): $0.01 \, J \, s^{-1} \, m^{-1} \, K^{-1}$
• Latent heat of fusion of water ($L_f$): $335 \times 10^3 \, J \, kg^{-1}$

Step 1: Calculating Surface Area

A cube has 6 faces. The surface area ($A$) of each face is $L^2$. Therefore, the total surface area for heat transfer is:

$A = 6 \times L^2 = 6 \times (0.30 \, m)^2 = 6 \times 0.09 \, m^2 = 0.54 \, m^2$

Step 2: Determining Total Heat Influx

Per the Fourier Law of Thermal Conduction: $Q = \frac{K \cdot A \cdot \Delta T \cdot t}{\Delta x}$

Substituting the values:

$Q = \frac{0.01 \cdot 0.54 \cdot 45 \cdot 21600}{0.05}$

$Q = \frac{0.243 \times 21600}{0.05} = \frac{5248.8}{0.05} = 104976 \, J$

Step 3: Calculating Mass of Melted Ice

The heat required to melt a mass $m$ of ice is $Q = m \cdot L_f$. Thus, $m = \frac{Q}{L_f}$

$m = \frac{104976}{335 \times 10^3} \, kg$

$m \approx 0.313 \, kg$

Step 4: Final Mass Remaining

$M_{remaining} = M_{initial} - m$

$M_{remaining} = 4.0 - 0.313 = 3.687 \, kg$

Final Solution: The amount of ice remaining after 6 hours is approximately 3.69 kg.

1. Theoretical Foundation: The Greenhouse Effect and Radiative Equilibrium

The Earth's surface temperature is determined by a radiative balance between incoming solar radiation (primarily short-wave visible light) and outgoing terrestrial radiation (long-wave infrared radiation). The atmosphere acts as a selective filter, governed by the following physical principles:

• Stefan-Boltzmann Law: The power radiated by a black body per unit area is proportional to the fourth power of its absolute temperature: $P = \sigma A T^4$ [Where $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$].
• Wien’s Displacement Law: The wavelength of maximum emission is inversely proportional to temperature: $\lambda_{max} T = b$ [Where $b \approx 2.898 \times 10^{-3} \, m \cdot K$].

2. Interaction of Solar and Terrestrial Radiation

3. Visualizing the Radiative Trap

4. Step-by-Step Analytical Breakdown

Step 1: The Without-Atmosphere Scenario
If the atmosphere were removed, the Earth would function as a bare black body in the vacuum of space. It would absorb solar radiation and emit thermal radiation directly into space without any re-absorption. The effective temperature ($T_e$) would be approximately $255 \, K$ ($-18^\circ C$).

Step 2: The Mechanism of Atmospheric Retention
Greenhouse gases in the atmosphere possess vibrational/rotational energy states that correspond to the frequencies of outgoing infrared (IR) radiation. By absorbing these photons, the atmosphere becomes heated [Per the Principle of Selective Absorption].

Step 3: Back-Radiation
The warmed atmosphere re-radiates a significant portion of this thermal energy back toward the Earth's surface. This "Back-Radiation" adds to the energy flux received from the Sun, raising the equilibrium temperature of the surface to the current global average of approximately $288 \, K$ ($15^\circ C$).

5. Conclusion

Without the atmosphere, the lack of this "radiative insulation" would result in a massive net loss of energy to space. The Earth would not be able to maintain the thermal energy required to sustain liquid water or biological processes, leading to an average surface temperature drop of roughly $33^\circ C$, rendering the planet inhospitable.

Final Solution: The atmosphere traps outgoing infrared radiation through the greenhouse effect, maintaining a surface temperature higher than the radiative equilibrium of a vacuum-exposed Earth; without this insulation, the Earth would radiate heat rapidly into space, resulting in sub-zero temperatures across most of the planet.

Given Variables & Initial Setup

• Calibration temperature of tape ($T_0$): $27.0 \, ^\circ\text{C}$
• Temperature on the hot day ($T$): $45.0 \, ^\circ\text{C}$
• Observed length of the rod at $T$ ($L'_{obs}$): $63.0 \, \text{cm}$
• Coefficient of linear expansion of steel ($\alpha$): $1.20 \times 10^{-5} \, \text{K}^{-1}$

Step 1: Calculating the Thermal Expansion of the Tape

The tape expands as the temperature increases from $27.0 \, ^\circ\text{C}$ to $45.0 \, ^\circ\text{C}$. The change in temperature is $\Delta T = 45.0 - 27.0 = 18.0 \, \text{K}$.

The actual length ($L$) of a scale marking representing $1 \, \text{cm}$ on the tape at the hot day is given by the linear expansion formula [Per the Thermal Expansion Law]:

$L = L_0(1 + \alpha \Delta T)$

$L = 1 \, \text{cm} \times [1 + (1.20 \times 10^{-5} \, \text{K}^{-1})(18.0 \, \text{K})]$

$L = 1 + 2.16 \times 10^{-4} = 1.000216 \, \text{cm}$

Step 2: Determining the Actual Length of the Rod at 45.0 °C

The rod length measured by the tape is $63.0$ markings. Since each marking corresponds to $1.000216 \, \text{cm}$ at $45.0 \, ^\circ\text{C}$, the actual length ($L_{actual, 45}$) is:

$L_{actual, 45} = 63.0 \times 1.000216 \, \text{cm}$

$L_{actual, 45} = 63.013608 \, \text{cm} \approx 63.014 \, \text{cm}$

Step 3: Calculating the Length of the Rod at 27.0 °C

To find the length of the rod at the calibration temperature ($27.0 \, ^\circ\text{C}$), we reverse the thermal expansion of the rod itself using the same coefficient [Per the Law of Thermal Contraction]:

$L_{27} = \frac{L_{actual, 45}}{1 + \alpha \Delta T}$

$L_{27} = \frac{63.013608}{1 + (1.20 \times 10^{-5})(18.0)}$

$L_{27} = \frac{63.013608}{1.000216} = 63.0 \, \text{cm}$

Summary of Results

Final Solution: The actual length of the rod at 45.0 °C is 63.014 cm, and at 27.0 °C, it is 63.0 cm.

Given Variables & Initial Setup

The following parameters are provided for the substance glycerine:

• Coefficient of volume expansion ($\gamma$): $49 \times 10^{-5} \, \text{K}^{-1}$
• Change in temperature ($\Delta T$): $30 \, ^\circ\text{C} = 30 \, \text{K}$

[Per the definition of thermal expansion, the volume of a substance increases as temperature increases, and since mass remains constant, density must decrease.]

Step 1: Derivation of the Relationship between Density and Temperature

The density ($\rho$) of a substance is defined as mass ($m$) divided by volume ($V$):

$\rho = \frac{m}{V}$

For a small change in temperature $\Delta T$, the new volume $V'$ is given by the formula:

$V' = V(1 + \gamma \Delta T)$

The new density $\rho'$ is therefore:

$\rho' = \frac{m}{V'} = \frac{m}{V(1 + \gamma \Delta T)} = \rho(1 + \gamma \Delta T)^{-1}$

[Using the Binomial Theorem $(1 + x)^n \approx 1 + nx$ for small $x$, where $\gamma \Delta T \ll 1$]:

$\rho' \approx \rho(1 - \gamma \Delta T)$

Step 2: Calculating Fractional Change in Density

The fractional change in density is defined as the ratio of the change in density ($\Delta \rho$) to the original density ($\rho$):

$\Delta \rho = \rho' - \rho = \rho(1 - \gamma \Delta T) - \rho = -\rho \gamma \Delta T$

The fractional change is:

$\frac{\Delta \rho}{\rho} = -\gamma \Delta T$

[The negative sign indicates a decrease in density as temperature rises].

Step 3: Numerical Substitution

Substitute the given values into the derived equation:

$\frac{\Delta \rho}{\rho} = -(49 \times 10^{-5} \, \text{K}^{-1}) \times (30 \, \text{K})$

$\frac{\Delta \rho}{\rho} = -(49 \times 30) \times 10^{-5}$

$\frac{\Delta \rho}{\rho} = -1470 \times 10^{-5}$

$\frac{\Delta \rho}{\rho} = -1.47 \times 10^{-2}$

$\frac{\Delta \rho}{\rho} = -0.0147$

Final Solution: The fractional decrease in the density of glycerine for a 30 °C rise in temperature is $1.47 \times 10^{-2}$ (or approximately $1.47\%$).

Step 1: Given Variables & Initial Setup

The relationship between electrical resistance $R$ and temperature $T$ is governed by the linear model:

$R = R_0 [1 + \alpha (T - T_0)]$

From the problem statement, we identify the following data points:

• Reference Temperature ($T_0$): $273.16\text{ K}$ (Triple point of water)
• Resistance at $T_0$ ($R_0$): $101.6\text{ }\Omega$
• Temperature at melting point of lead ($T_1$): $600.5\text{ K}$
• Resistance at $T_1$ ($R_1$): $165.5\text{ }\Omega$
• Target Resistance ($R_t$): $123.4\text{ }\Omega$
• Target Temperature ($T$): To be determined.

Step 2: Determining the Temperature Coefficient ($\alpha$)

Using the linear equation for the known state at the melting point of lead, we isolate the coefficient $\alpha$:

$R_1 = R_0 [1 + \alpha(T_1 - T_0)]$

Substituting the known values:

$165.5 = 101.6 [1 + \alpha(600.5 - 273.16)]$

$165.5 = 101.6 [1 + \alpha(327.34)]$

Divide by $101.6$:

$1.628937 = 1 + \alpha(327.34)$

$0.628937 = \alpha(327.34)$

$\alpha = \frac{0.628937}{327.34} \approx 1.92135 \times 10^{-3}\text{ K}^{-1}$

Step 3: Calculating the Target Temperature ($T$)

We rearrange the original formula to solve for $T$ at $R_t = 123.4\text{ }\Omega$:

$R_t = R_0 [1 + \alpha(T - T_0)]$

$\frac{R_t}{R_0} = 1 + \alpha(T - T_0)$

$\frac{R_t}{R_0} - 1 = \alpha(T - T_0)$

$T = T_0 + \frac{R_t - R_0}{\alpha \cdot R_0}$

Substituting the values:

$T = 273.16 + \frac{123.4 - 101.6}{(1.92135 \times 10^{-3}) \times 101.6}$

$T = 273.16 + \frac{21.8}{0.195209}$

$T = 273.16 + 111.675$

$T \approx 384.835\text{ K}$

Visual Representation of Linear Resistance Model

Final Solution: The temperature when the resistance is 123.4 Ω is approximately 384.84 K.

Given Variables & Initial Setup

According to Newton’s Law of Cooling, the rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings, provided the difference is small. The formula is expressed as:

$$\frac{dT}{dt} = -K(T - T_s)$$

Where:

• $T$ = Temperature of the body
• $T_s$ = Temperature of the surroundings = $20^\circ\text{C}$
• $K$ = Positive constant
• $t$ = Time

For finite temperature differences, we use the integrated form (average temperature approximation):

$$\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_s\right)$$

Step 1: Analyzing the First Cooling Interval

Interval: $80^\circ\text{C}$ to $50^\circ\text{C}$ in $t_1 = 5$ minutes.

Substituting the values into the cooling formula:

$$\frac{80 - 50}{5} = K\left(\frac{80 + 50}{2} - 20\right)$$

$$\frac{30}{5} = K(65 - 20)$$

$$6 = K(45) \implies K = \frac{6}{45} = \frac{2}{15} \text{ min}^{-1}$$

[Justification: Per Newton's Law of Cooling applied to the interval mean temperature.]

Step 2: Analyzing the Second Cooling Interval

Interval: $60^\circ\text{C}$ to $30^\circ\text{C}$ in time $t_2$.

Using the constant $K$ derived in Step 1:

$$\frac{60 - 30}{t_2} = K\left(\frac{60 + 30}{2} - 20\right)$$

$$\frac{30}{t_2} = \left(\frac{2}{15}\right)(45 - 20)$$

$$\frac{30}{t_2} = \left(\frac{2}{15}\right)(25)$$

$$\frac{30}{t_2} = \frac{50}{15} = \frac{10}{3}$$

Step 3: Calculating Final Time

Solving for $t_2$:

$$t_2 = \frac{30 \times 3}{10}$$

$$t_2 = 9 \text{ minutes}$$

Visual Representation of Cooling Curve

Final Solution: The time taken to cool from 60 °C to 30 °C is 9 minutes.

Step 1: Identifying Given Variables

• Side of the cube ($l$): $30\text{ cm} = 0.3\text{ m}$
• Thickness of thermacole ($\Delta x$): $5.0\text{ cm} = 0.05\text{ m}$
• Mass of ice ($m_{initial}$): $4.0\text{ kg}$
• Time duration ($t$): $6\text{ h} = 6 \times 3600\text{ s} = 21,600\text{ s}$
• Temperature difference ($\Delta T$): $T_{outside} - T_{inside} = 45^\circ\text{C} - 0^\circ\text{C} = 45\text{ K}$
• Thermal conductivity ($K$): $0.01\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1}$
• Latent heat of fusion of ice ($L_f$): $335 \times 10^3\text{ J kg}^{-1}$

Step 2: Calculating Surface Area for Heat Transfer

A cube has 6 faces. The total surface area ($A$) is:

Step 3: Calculating Total Heat Conducted ($Q$)

Using the steady-state heat conduction formula [Fourier’s Law of Heat Conduction]:

Step 4: Calculating Mass of Ice Melted ($m_{melted}$)**