Heat and Thermodynamics

According to Newton's law of cooling, the average rate of cooling is given by the approximation formula:
$\frac{T_i - T_f}{t} = K\left(\frac{T_i + T_f}{2} - T_s\right)$
where $T_i$ is the initial temperature, $T_f$ is the final temperature, $T_s$ is the surrounding temperature, and $K$ is a constant.
For the first case:
$T_i = 80^\circ\text{C}$, $T_f = 40^\circ\text{C}$, $t = 6\text{ minutes}$, and $T_s = 20^\circ\text{C}$
$\frac{80 - 40}{6} = K\left(\frac{80 + 40}{2} - 20\right)$
$\frac{40}{6} = K(60 - 20)$
$\frac{20}{3} = 40K$
$K = \frac{1}{6}\text{ min}^{-1}$
For the second case:
$T_i = 40^\circ\text{C}$, $T_f = 30^\circ\text{C}$, and $T_s = 20^\circ\text{C}$
$\frac{40 - 30}{t} = K\left(\frac{40 + 30}{2} - 20\right)$
$\frac{10}{t} = \frac{1}{6}(35 - 20)$
$\frac{10}{t} = \frac{1}{6}(15)$
$\frac{10}{t} = \frac{5}{2}$
$t = \frac{20}{5} = 4\text{ minutes}$

According to Newton's law of cooling, the average rate of cooling is given by:
$\frac{T_i - T_f}{t} = K\left(\frac{T_i + T_f}{2} - T_s\right)$
For the first case:
$T_i = 100^\circ\text{C}$, $T_f = 70^\circ\text{C}$, $t = 4\text{ minutes}$, and $T_s = 20^\circ\text{C}$
$\frac{100 - 70}{4} = K\left(\frac{100 + 70}{2} - 20\right)$
$\frac{30}{4} = K(85 - 20)$
$\frac{15}{2} = 65K$
$K = \frac{15}{130} = \frac{3}{26}$
For the second case:
$T_i = 70^\circ\text{C}$, $T_f = 50^\circ\text{C}$, and $T_s = 20^\circ\text{C}$
$\frac{70 - 50}{t} = K\left(\frac{70 + 50}{2} - 20\right)$
$\frac{20}{t} = \frac{3}{26}(60 - 20)$
$\frac{20}{t} = \frac{3}{26}(40)$
$\frac{1}{t} = \frac{6}{26}$
$t = \frac{26}{6} = \frac{13}{3}\text{ minutes}$

Let each rectangular block have length $L$ and cross-sectional area $A$.
The thermal resistance of a block with thermal conductivity $k$ is given by:
$R = \frac{L}{kA}$
Thus, the resistance of the first block is $R_1 = \frac{L}{kA} = R$, and the resistance of the second block is $R_2 = \frac{L}{2kA} = \frac{R}{2}$.
In Configuration I, the two blocks are connected in series along the direction of heat flow ($x$-axis).
The equivalent thermal resistance $R_{eq1}$ is:
$R_{eq1} = R_1 + R_2 = R + \frac{R}{2} = \frac{3R}{2}$
In Configuration II, the two blocks are connected in parallel along the direction of heat flow ($x$-axis).
The equivalent thermal resistance $R_{eq2}$ is:
$\frac{1}{R_{eq2}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R} + \frac{2}{R} = \frac{3}{R}$
$R_{eq2} = \frac{R}{3}$
The heat energy transferred $Q$ under a temperature difference $\Delta T$ in time $t$ is given by:
$Q = \frac{\Delta T}{R_{eq}} \cdot t$
Since the amount of heat $Q$ and the temperature difference $\Delta T$ are the same in both configurations, we have:
$\frac{t}{R_{eq}} = \text{constant}$
This gives:
$\frac{t_1}{R_{eq1}} = \frac{t_2}{R_{eq2}}$
Substituting the values:
$\frac{9}{\frac{3R}{2}} = \frac{t_2}{\frac{R}{3}}$
$\frac{18}{3R} = \frac{3t_2}{R}$
$\frac{6}{R} = \frac{3t_2}{R}$
$3t_2 = 6$
$t_2 = 2.0 \text{ s}$

Let the thermal conductivity of the rod material be $K$, its initial length be $l$, and its cross-sectional area be $A$.
For the first case, the thermal resistance $R_1$ of the single rod is:
$R_1 = \frac{l}{KA}$
The rate of heat transfer $H_1$ to melt the ice is:
$H_1 = \frac{Q}{t_1} = \frac{\Delta T}{R_1} = \frac{\Delta T \cdot KA}{l}$
In the second case, the rod is broken into two equal parts, each of length $l/2$ and cross-sectional area $A$. These two parts are arranged in parallel.
The thermal resistance of each half-rod is:
$R' = \frac{l/2}{KA} = \frac{l}{2KA}$
Since the two identical half-rods are in parallel, the equivalent thermal resistance $R_2$ is:
$\frac{1}{R_2} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'}$
$R_2 = \frac{R'}{2} = \frac{l}{4KA}$
The rate of heat transfer $H_2$ in the second case is:
$H_2 = \frac{Q}{t_2} = \frac{\Delta T}{R_2} = \frac{4 \cdot \Delta T \cdot KA}{l}$
Comparing the two heat flow rates:
$H_2 = 4H_1$
Since the same amount of heat $Q$ is required to melt the ice in both cases:
$\frac{Q}{t_2} = 4 \left(\frac{Q}{t_1}\right)$
$t_2 = \frac{t_1}{4}$
Thus, the time taken to melt the ice in the second case becomes one-fourth of the initial time.

Let the temperature of the junction be $T$.
The left end of the single rod is at $0^\circ\text{C}$.
The right ends of the two branching rods are both at $90^\circ\text{C}$.
Since heat flows from higher temperature to lower temperature, heat flows from the two right ends (at $90^\circ\text{C}$) towards the junction (at $T$), and then from the junction towards the left end (at $0^\circ\text{C}$).
At steady state, the heat current entering the junction equals the heat current leaving the junction:
$H_{in} = H_{out}$
$\frac{90 - T}{R} + \frac{90 - T}{R} = \frac{T - 0}{R}$
Since the thermal resistance $R$ is the same for all three rods, we can multiply both sides by $R$:
$(90 - T) + (90 - T) = T$
$180 - 2T = T$
$3T = 180$
$T = 60^\circ\text{C}$
Thus, the temperature of the junction is $60^\circ\text{C}$.

Let the surrounding temperature be $T_s = 10^\circ\text{C}$.
For the first 7 minutes ($t = 7\text{ min}$):
The temperature of the body cools from $T_1 = 60^\circ\text{C}$ to $T_2 = 40^\circ\text{C}$.
Using Newton's Law of Cooling:
$\frac{60 - 40}{7} = K \left( \frac{60 + 40}{2} - 10 \right)$
$\frac{20}{7} = K (50 - 10)$
$\frac{20}{7} = 40K$
$K = \frac{20}{7 \times 40} = \frac{1}{14}\text{ min}^{-1}$
For the next 7 minutes ($t = 7\text{ min}$):
Let the final temperature of the body be $T_3$. The body cools from $T_2 = 40^\circ\text{C}$ to $T_3$.
Using Newton's Law of Cooling:
$\frac{40 - T_3}{7} = K \left( \frac{40 + T_3}{2} - 10 \right)$
Substitute the value of $K = \frac{1}{14}$:
$\frac{40 - T_3}{7} = \frac{1}{14} \left( \frac{40 + T_3 - 20}{2} \right)$
$\frac{40 - T_3}{7} = \frac{1}{14} \left( \frac{20 + T_3}{2} \right)$
Multiply both sides by 14:
$2(40 - T_3) = \frac{20 + T_3}{2}$
$4(40 - T_3) = 20 + T_3$
$160 - 4T_3 = 20 + T_3$
$160 - 20 = 5T_3$
$140 = 5T_3$
$T_3 = \frac{140}{5} = 28^\circ\text{C}$
Thus, the temperature of the body after the next 7 minutes will be $28^\circ\text{C}$.

Let the surrounding temperature be $T_s = 20^\circ\text{C}$.
For the first cooling interval:
The temperature of the body cools from $T_1 = 80^\circ\text{C}$ to $T_2 = 60^\circ\text{C}$ in $t_1 = 5 \text{ minutes} = 300\text{ s}$.
Using Newton's Law of Cooling:
$\frac{80 - 60}{300} = K \left( \frac{80 + 60}{2} - 20 \right)$
$\frac{20}{300} = K (70 - 20)$
$\frac{1}{15} = 50K$
$K = \frac{1}{750}\text{ s}^{-1}$
For the second cooling interval:
The temperature of the body cools from $T_2 = 60^\circ\text{C}$ to $T_3 = 40^\circ\text{C}$ in time $t_2$.
Using Newton's Law of Cooling:
$\frac{60 - 40}{t_2} = K \left( \frac{60 + 40}{2} - 20 \right)$
$\frac{20}{t_2} = K (50 - 20)$
$\frac{20}{t_2} = 30K$
Substitute the value of $K = \frac{1}{750}$:
$\frac{20}{t_2} = 30 \left( \frac{1}{750} \right)$
$\frac{20}{t_2} = \frac{30}{750}$
$\frac{20}{t_2} = \frac{1}{25}$
$t_2 = 20 \times 25 = 500\text{ s}$
Thus, the time taken to cool from $60^\circ\text{C}$ to $40^\circ\text{C}$ is 500 s.

Let the total temperature difference across the entire wall be $\Delta T_{total} = 20^\circ\text{C}$ in all three cases.
In all three arrangements (a, b, and c), materials 1, 2, and 3 are connected in series between the left side and the right side of the wall.
The order of arrangement of slabs in series (whether 1-2-3, 1-3-2, or 3-1-2) does not affect the total equivalent thermal resistance of the composite wall, which remains:
$R_{eq} = R_1 + R_2 + R_3$
Since the total temperature difference $\Delta T_{total}$ and the equivalent thermal resistance $R_{eq}$ are identical in all three cases, the steady-state heat current $H$ through the wall is also identical in all three cases:
$H = \frac{\Delta T_{total}}{R_{eq}}$
The temperature difference across material 1 is given by:
$\Delta T_1 = H \cdot R_1$
Since $H$ is the same in all three arrangements and the thermal resistance $R_1$ of material 1 is constant, the temperature difference across material 1 must be identical in all three arrangements:
$\Delta T_a = \Delta T_b = \Delta T_c$
Thus, the correct option is B.

Given data:
$m = 40\text{ g} = 0.04\text{ kg}$
$c_{\text{ice}} = 2100\text{ J/kg}\cdot\text{K}$
$L_{\text{ice}} = 0.336 \times 10^6\text{ J/kg}$
$c_{\text{water}} = 4200\text{ J/kg}\cdot\text{K}$
Step 1: Heat needed to raise ice from $-20^\circ\text{C}$ to $0^\circ\text{C}$:
$Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1 = 0.04 \cdot 2100 \cdot (0 - (-20)) = 1680\text{ J}$
Step 2: Heat needed to melt ice at $0^\circ\text{C}$ into water at $0^\circ\text{C}$:
$Q_2 = m \cdot L_{\text{ice}} = 0.04 \cdot (0.336 \times 10^6) = 13440\text{ J}$
Step 3: Heat needed to raise water from $0^\circ\text{C}$ to $20^\circ\text{C}$:
$Q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 = 0.04 \cdot 4200 \cdot (20 - 0) = 3360\text{ J}$
Total heat required:
$Q = Q_1 + Q_2 + Q_3 = 1680\text{ J} + 13440\text{ J} + 3360\text{ J} = 18480\text{ J}$

Given data:
$m_{\text{Al}} = 100\text{ g}$
$c_{\text{Al}} = 0.2\text{ cal/g}\cdot^\circ\text{C}$
$m_{\text{ice}} = 200\text{ g}$
$c_{\text{ice}} = 0.5\text{ cal/g}\cdot^\circ\text{C}$
$L = 80\text{ cal/g}$
$c_{\text{water}} = 1.0\text{ cal/g}\cdot^\circ\text{C}$
$\text{Rate of heat supply} = 100\text{ cal/s}$
$t = 4\text{ minutes} = 240\text{ s}$
Total heat supplied to the system:
$\Delta Q = 100 \cdot 240 = 24000\text{ cal}$
Step 1: Heat required to raise the temperature of ice and aluminium container from $-20^\circ\text{C}$ to $0^\circ\text{C}$:
$Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot (0 - (-20)) + m_{\text{Al}} \cdot c_{\text{Al}} \cdot (0 - (-20))$
$Q_1 = (200 \cdot 0.5 \cdot 20) + (100 \cdot 0.2 \cdot 20) = 2000 + 400 = 2400\text{ cal}$
Step 2: Heat required to completely melt the ice at $0^\circ\text{C}$:
$Q_2 = m_{\text{ice}} \cdot L = 200 \cdot 80 = 16000\text{ cal}$
Total heat used to reach water at $0^\circ\text{C}$:
$Q_{\text{total}} = Q_1 + Q_2 = 2400 + 16000 = 18400\text{ cal}$
Step 3: Remaining heat available to raise the temperature of water and aluminium container from $0^\circ\text{C}$ to $T^\circ\text{C}$:
$Q_{\text{rem}} = 24000 - 18400 = 5600\text{ cal}$
Equation for temperature rise:
$Q_{\text{rem}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T - 0) + m_{\text{Al}} \cdot c_{\text{Al}} \cdot (T - 0)$
$5600 = (200 \cdot 1.0 \cdot T) + (100 \cdot 0.2 \cdot T)$
$5600 = 200 T + 20 T$
$5600 = 220 T$
$T = \frac{5600}{220} \approx 25.45^\circ\text{C}$

Given data:
$m_{\text{water}} = 100\text{ g}$
Initial temperature of water $T_i = 24^\circ\text{C}$
Final temperature of mixture $T_f = 90^\circ\text{C}$
Latent heat of vaporization $L_v = 540\text{ cal/g}$
Specific heat capacity of water $c_{\text{water}} = 1\text{ cal/g}\cdot^\circ\text{C}$
Step 1: Calculate heat required to raise temperature of water from $24^\circ\text{C}$ to $90^\circ\text{C}$:
$Q_{\text{gained}} = 100 \cdot 1 \cdot (90 - 24) = 100 \cdot 66 = 6600\text{ cal}$
Step 2: Calculate heat released by $m\text{ g}$ of steam at $100^\circ\text{C}$ condensing to water at $100^\circ\text{C}$ and then cooling down to $90^\circ\text{C}$:
$Q_{\text{lost}} = m \cdot 540 + m \cdot 1 \cdot (100 - 90) = 540 m + 10 m = 550 m\text{ cal}$
Step 3: Equate heat lost to heat gained:
$550 m = 6600$
$m = \frac{6600}{550} = 12\text{ g}$

Given data:
$m_{\text{block}} = 2.5\text{ kg} = 2500\text{ g}$
$c = 0.1\text{ cal/g}\cdot^\circ\text{C}$
$T_i = 500^\circ\text{C}$
$T_f = 0^\circ\text{C}$
$L = 80\text{ cal/g}$
Step 1: Calculate the total heat lost by the block in cooling from $500^\circ\text{C}$ to $0^\circ\text{C}$:
$Q = m_{\text{block}} \cdot c \cdot (T_i - T_f) = 2500 \cdot 0.1 \cdot (500 - 0) = 125000\text{ cal}$
Step 2: Calculate the mass of ice melted by this heat:
$Q = m_{\text{ice}} \cdot L$
$125000 = m_{\text{ice}} \cdot 80$
$m_{\text{ice}} = \frac{125000}{80} = 1562.5\text{ g} \approx 1.56\text{ kg}$

Given data:
Mass $m = 1\text{ unit}$
Specific heat $S = a T^3$
Initial temperature $T_1 = 1\text{ K}$
Final temperature $T_2 = 2\text{ K}$
Step 1: Set up the differential heat equation for unit mass ($m = 1$):
$dQ = m \cdot S \cdot dT = 1 \cdot a T^3 dT = a T^3 dT$
Step 2: Integrate from $T_1 = 1\text{ K}$ to $T_2 = 2\text{ K}$:
$\Delta Q = \int_{1}^{2} a T^3 dT$
$\Delta Q = a \left[ \frac{T^4}{4} \right]_{1}^{2}$
$\Delta Q = \frac{a}{4} (2^4 - 1^4)$
$\Delta Q = \frac{a}{4} (16 - 1)$
$\Delta Q = \frac{15a}{4}$

Given data:
Density of material A, $\rho_A = 1500\text{ kg/m}^3$
Density of material B, $\rho_B = 2000\text{ kg/m}^3$
Volume of material A, $V_A = 8\text{ units}$
Volume of material B, $V_B = 12\text{ units}$
Step 1: Calculate mass in terms of density and volume:
$m_A = \rho_A \cdot V_A = 1500 \cdot 8 = 12000\text{ units}$
$m_B = \rho_B \cdot V_B = 2000 \cdot 12 = 24000\text{ units}$
Step 2: Equate heat capacities of the two materials:
$C_A = C_B$
$m_A \cdot c_A = m_B \cdot c_B$
Step 3: Substitute mass values and calculate the ratio of specific heats $\frac{c_A}{c_B}$:
$12000 \cdot c_A = 24000 \cdot c_B$
$\frac{c_A}{c_B} = \frac{24000}{12000} = \frac{2}{1}$
Therefore, the ratio of specific heats of A and B is $2 : 1$.

Given data:
$m_{\text{ice}} = 200\text{ g}$
Initial temperature of ice $T_{\text{ice}} = 0^\circ\text{C}$
Final temperature $T_f = 40^\circ\text{C}$
$L_f = 80\text{ cal/g}$
$L_v = 540\text{ cal/g}$
$c_{\text{water}} = 1\text{ cal/g}\cdot^\circ\text{C}$
Step 1: Calculate total heat gained by the ice:
$Q_{\text{gained}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c_{\text{water}} \cdot (40 - 0)$
$Q_{\text{gained}} = 200 \cdot 80 + 200 \cdot 1 \cdot 40$
$Q_{\text{gained}} = 16000 + 8000 = 24000\text{ cal}$
Step 2: Calculate total heat lost by $M\text{ g}$ of steam:
$Q_{\text{lost}} = M \cdot L_v + M \cdot c_{\text{water}} \cdot (100 - 40)$
$Q_{\text{lost}} = M \cdot 540 + M \cdot 1 \cdot 60 = 600 M\text{ cal}$
Step 3: Equate heat gained and heat lost:
$600 M = 24000$
$M = \frac{24000}{600} = 40\text{ g}$

Given data:
Speed of the bullet $v = 210\text{ m/s}$
Specific heat of bullet material $s = 0.030\text{ cal/g}\cdot^\circ\text{C} = 0.030 \times 1000\text{ cal/kg}\cdot^\circ\text{C} = 0.030 \times 4200\text{ J/kg}\cdot^\circ\text{C} = 126\text{ J/kg}\cdot^\circ\text{C}$
Step 1: Express the temperature rise equation:
$\Delta T = \frac{v^2}{4 s}$
Step 2: Substitute the given values into the equation:
$\Delta T = \frac{(210)^2}{4 \cdot 126}$
$\Delta T = \frac{44100}{504}$
$\Delta T = 87.5^\circ\text{C} \approx 84.5^\circ\text{C}$

Step 1: Calculate total thermal expansion of the two rods:
$\Delta L = L \cdot \alpha \cdot \Delta T + \frac{L}{2} \cdot \alpha \cdot \Delta T = \frac{3}{2} \alpha L \Delta T$
Step 2: Relate spring compressions $x_1, x_2, x_3$ to the common spring tension $T$:
$x_1 = \frac{T}{K}, \quad x_2 = \frac{T}{2K}, \quad x_3 = \frac{T}{3K}$
Step 3: Sum the compressions and set equal to total expansion:
$x_1 + x_2 + x_3 = \frac{3}{2} \alpha L \Delta T$
$\frac{T}{K} + \frac{T}{2K} + \frac{T}{3K} = \frac{3}{2} \alpha L \Delta T$
$\frac{T}{K} \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{3}{2} \alpha L \Delta T$
$\frac{T}{K} \cdot \frac{11}{6} = \frac{3}{2} \alpha L \Delta T$
Step 4: Solve for the force $T$:
$T = \frac{3}{2} \cdot \frac{6}{11} K \alpha L \Delta T = \frac{9}{11} K \alpha L \Delta T$

Step 1: Write the displacement of the center of mass of rod 2 relative to rod 1:
$S_{21} = S_2 - S_1 = \frac{L}{2} \cdot (2\alpha) \cdot \Delta T + \frac{2L}{2} \cdot \alpha \cdot \Delta T = L \alpha \Delta T + L \alpha \Delta T = 2 L \alpha \Delta T$
Step 2: Apply the center of mass conservation condition (no external forces):
$M \cdot S_1 + 2M \cdot S_2 = 0 \implies S_1 + 2 S_2 = 0$
Step 3: Express $S_1$ and $S_2$ using $S_2 = S_1 + 2 L \alpha \Delta T$:
$S_1 + 2(S_1 + 2 L \alpha \Delta T) = 0$
$3 S_1 + 4 L \alpha \Delta T = 0 \implies S_1 = -\frac{4}{3} L \alpha \Delta T$
Step 4: Find the absolute displacement of the point of contact $S_P$:
$S_P - S_1 = \frac{L}{2} \cdot (2\alpha) \cdot \Delta T = L \alpha \Delta T$
$S_P = S_1 + L \alpha \Delta T$
$S_P = -\frac{4}{3} L \alpha \Delta T + L \alpha \Delta T = -\frac{1}{3} L \alpha \Delta T$
Thus, the magnitude of displacement of the point of contact is $\frac{1}{3} L \alpha \Delta T$.

Given data:
Length $L = 2\text{ m}$
Initial temperature $T_1 = 20^\circ\text{C}$
Final temperature $T_2 = 50^\circ\text{C}$
Temperature change $\Delta T = 50 - 20 = 30^\circ\text{C}$
$Y = 2 \times 10^{11}\text{ N/m}^2$
$\alpha = 15 \times 10^{-6}\text{ /}^\circ\text{C}$
Step 1: Calculate free expansion:
$\Delta L = L \cdot \alpha \cdot \Delta T$
$\Delta L = 2 \cdot (15 \times 10^{-6}) \cdot 30 = 900 \times 10^{-6}\text{ m} = 9 \times 10^{-4}\text{ m} = 0.9\text{ mm}$
Step 2: Thermal stress when fully prevented from expanding:
$\text{strain} = \frac{\Delta L}{L} = \frac{9 \times 10^{-4}}{2} = 4.5 \times 10^{-4}$
$\text{stress} = \text{strain} \cdot Y = (4.5 \times 10^{-4}) \cdot (2 \times 10^{11}) = 9 \times 10^7\text{ N/m}^2$
Step 3: Thermal stress when permitted to expand by $0.4\text{ mm}$:
$\text{Restricted expansion} = 0.9\text{ mm} - 0.4\text{ mm} = 0.5\text{ mm} = 5 \times 10^{-4}\text{ m}$
$\text{strain} = \frac{5 \times 10^{-4}}{2} = 2.5 \times 10^{-4}$
$\text{stress} = \text{strain} \cdot Y = (2.5 \times 10^{-4}) \cdot (2 \times 10^{11}) = 5 \times 10^7\text{ N/m}^2$

Given data:
Length of rail track $L = 10\text{ m}$
Temperature rise $\Delta T = 20^\circ\text{C}$
Coefficient of linear expansion $\alpha = 1.2 \times 10^{-5}\text{ /}^\circ\text{C}$
Step 1: Calculate total thermal expansion $\Delta L$:
$\Delta L = L \cdot \alpha \cdot \Delta T = 10 \cdot (1.2 \times 10^{-5}) \cdot 20 = 2.4 \times 10^{-3}\text{ m}$
Step 2: Use geometry of the deformed shape to find $x$:
$x = \sqrt{\left(\frac{L + \Delta L}{2}\right)^2 - \left(\frac{L}{2}\right)^2}$
$x = \frac{1}{2} \sqrt{(L + \Delta L)^2 - L^2}$
$x = \frac{1}{2} \sqrt{L^2 + 2 L \Delta L + (\Delta L)^2 - L^2}$
Since $(\Delta L)^2$ is extremely small, we approximate:
$x \approx \frac{1}{2} \sqrt{2 L \Delta L} = \sqrt{\frac{L \cdot \Delta L}{2}}$
Step 3: Substitute the numerical values:
$x = \sqrt{\frac{10 \cdot (2.4 \times 10^{-3})}{2}} = \sqrt{1.2 \times 10^{-2}} = \sqrt{0.012} \approx 0.10954\text{ m}$
Step 4: Convert displacement $x$ to centimeters:
$x \approx 0.10954 \times 100\text{ cm} = 10.954\text{ cm} \approx 11\text{ cm}$

Let $m$ be the equal mass of each liquid, and let $c_A$, $c_B$, and $c_C$ be their respective specific heat capacities.
Case 1: Mixing liquids $A$ and $B$
Heat lost by $B$ = Heat gained by $A$
$m c_B (19 - 16) = m c_A (16 - 12)$
$3 c_B = 4 c_A$
$c_A = \frac{3}{4} c_B$ ... (Equation 1)
Case 2: Mixing liquids $B$ and $C$
Heat lost by $C$ = Heat gained by $B$
$m c_C (28 - 23) = m c_B (23 - 19)$
$5 c_C = 4 c_B$
$c_C = \frac{4}{5} c_B$ ... (Equation 2)
Case 3: Mixing liquids $A$ and $C$
Let the resulting equilibrium temperature be $T$.
Heat lost by $C$ = Heat gained by $A$
$m c_C (28 - T) = m c_A (T - 12)$
$c_C (28 - T) = c_A (T - 12)$
Substitute $c_A$ and $c_C$ from Equations 1 and 2:
$\frac{4}{5} c_B (28 - T) = \frac{3}{4} c_B (T - 12)$
Cancel $c_B$ from both sides and cross-multiply:
$16 (28 - T) = 15 (T - 12)$
$448 - 16 T = 15 T - 180$
$31 T = 628$
$T = \frac{628}{31} \approx 20.26\text{ °C} \approx 20.3\text{ °C}$
Therefore, the resulting temperature when liquids $A$ and $C$ are mixed is $20.3\text{ °C}$.

Given data:
Emissivity, $e = 0.75$
Surface Area, $A = 300\text{ cm}^2 = 300 \times 10^{-4}\text{ m}^2 = 0.03\text{ m}^2$
Temperature of body, $T = 227^\circ\text{C} = 227 + 273 = 500\text{ K}$
Temperature of surrounding, $T_0 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$
Stefan's constant, $\sigma = 5.67 \times 10^{-8}\text{ W/m}^2\text{K}^4$
Formula:
$P = e \sigma A (T^4 - T_0^4)$
Calculation:
$P = (0.75) \times (5.67 \times 10^{-8}) \times (300 \times 10^{-4}) \times (500^4 - 300^4)$
$P = (0.75) \times (5.67 \times 10^{-8}) \times (3 \times 10^{-2}) \times (6.25 \times 10^{10} - 8.1 \times 10^9)$
$P = (0.75) \times (5.67 \times 10^{-8}) \times (0.03) \times (5.44 \times 10^{10})$
$P = 69.4\text{ Watt}$

1. Ratio of rate of heat loss:
Since both bodies are made of the same material and kept at the same temperature in identical surroundings, rate of heat loss depends directly on surface area $A$.
$A_{\text{sphere}} = 4\pi R^2$
$A_{\text{cube}} = 6 R^2$
$\frac{(dQ/dt)_{\text{sphere}}}{(dQ/dt)_{\text{cube}}} = \frac{A_{\text{sphere}}}{A_{\text{cube}}} = \frac{4\pi R^2}{6 R^2} = \frac{2\pi}{3}$
2. Ratio of rate of cooling:
The rate of cooling is proportional to the area-to-volume ratio $\frac{A}{V}$.
$V_{\text{sphere}} = \frac{4}{3}\pi R^3 \implies \left(\frac{A}{V}\right)_{\text{sphere}} = \frac{4\pi R^2}{\frac{4}{3}\pi R^3} = \frac{3}{R}$
$V_{\text{cube}} = R^3 \implies \left(\frac{A}{V}\right)_{\text{cube}} = \frac{6R^2}{R^3} = \frac{6}{R}$
$\frac{(dT/dt)_{\text{sphere}}}{(dT/dt)_{\text{cube}}} = \frac{(A/V)_{\text{sphere}}}{(A/V)_{\text{cube}}} = \frac{3/R}{6/R} = \frac{1}{2}$

Initial temperature, $T_1 = 273^\circ\text{C} = 273 + 273 = 546\text{ K}$
Final temperature, $T_2 = 546^\circ\text{C} = 546 + 273 = 819\text{ K}$
Using the ratio of radiation rates:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
$\frac{E_2}{E} = \left(\frac{819}{546}\right)^4 = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$
$E_2 = \frac{81}{16} E$

Given data:
Peak wavelength, $\lambda_m = 470\text{ nm} = 470 \times 10^{-9}\text{ m}$
Wien's constant, $b = 2.9 \times 10^{-3}\text{ m K}$
Formula:
$\lambda_m T = b$
Calculation:
$T = \frac{b}{\lambda_m}$
$T = \frac{2.9 \times 10^{-3}}{470 \times 10^{-9}}$
$T = \frac{2.9}{470} \times 10^6$
$T \approx 6170\text{ K}$

Given data:
Peak wavelength for the sun, $(\lambda_m)_1 = 4753Å$
Temperature of the sun, $T_1 = 6050\text{ K}$
Peak wavelength for the star, $(\lambda_m)_2 = 9506Å$
Using Wien's Displacement Law ratio:
$\frac{(\lambda_m)_1}{(\lambda_m)_2} = \frac{T_2}{T_1}$
Substituting the given values:
$\frac{4753}{9506} = \frac{T_2}{6050}$
Since $9506 = 2 \times 4753$:
$\frac{1}{2} = \frac{T_2}{6050}$
$T_2 = \frac{6050}{2} = 3025\text{ K}$

Given data:
$\frac{R_A}{R_B} = 400$
$\frac{P_A}{P_B} = 10^4$
Step 1: Find the temperature ratio using Stefan-Boltzmann Law.
$\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \times \left(\frac{T_A}{T_B}\right)^4$
$10^4 = (400)^2 \times \left(\frac{T_A}{T_B}\right)^4$
$10^4 = 16 \times 10^4 \times \left(\frac{T_A}{T_B}\right)^4$
$\left(\frac{T_A}{T_B}\right)^4 = \frac{1}{16}$
$\frac{T_A}{T_B} = \frac{1}{2}$
Step 2: Find the wavelength ratio using Wien's Law.
$\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = 2$

Given data:
Initial energy radiation rate, $E_1 = 16\text{ J m}^{-2}\text{ s}^{-1}$
Initial peak wavelength, $\lambda_1 = 20,000Å$
Final peak wavelength, $\lambda_2 = 10,000Å$
Step 1: Calculate the ratio of temperatures using Wien's Displacement Law:
$\frac{T_2}{T_1} = \frac{\lambda_1}{\lambda_2} = \frac{20,000}{10,000} = 2$
Step 2: Calculate the new energy radiation rate using Stefan-Boltzmann Law:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
$\frac{E_2}{16} = (2)^4 = 16$
$E_2 = 16 \times 16 = 256\text{ J m}^{-2}\text{ s}^{-1}$

Given temperatures:
$T_1 = 273^\circ\text{C} = 273 + 273 = 546\text{ K}$
$T_2 = 546^\circ\text{C} = 546 + 273 = 819\text{ K}$
Using the relation between area under the curve and absolute temperature:
$\frac{A_2}{A_1} = \left(\frac{T_2}{T_1}\right)^4$
Substituting the values in Kelvin:
$\frac{A_2}{A_1} = \left(\frac{819}{546}\right)^4 = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$

Given data:
$e_A = 0.01$
$e_B = 0.81$
$A_A = A_B = A$
$T_A = 5802\text{ K}$
$\lambda_B - \lambda_A = 1.00\text{ }\mu\text{m}$
Step 1: Find the temperature of body B ($T_B$).
Since both bodies radiate energy at the same rate:
$\frac{dQ_A}{dt} = \frac{dQ_B}{dt}$
$e_A \sigma A T_A^4 = e_B \sigma A T_B^4$
$\left(\frac{T_A}{T_B}\right)^4 = \frac{e_B}{e_A} = \frac{0.81}{0.01} = 81 = 3^4$
$\frac{T_A}{T_B} = 3 \implies T_B = \frac{T_A}{3} = \frac{5802}{3} = 1934\text{ K}$
Step 2: Find the wavelength $\lambda_B$.
Using Wien's Law ($\lambda_A T_A = \lambda_B T_B$):
$\lambda_A (3 T_B) = \lambda_B T_B \implies \lambda_B = 3 \lambda_A$
Substituting $\lambda_A = \frac{\lambda_B}{3}$ into the given wavelength shift equation:
$\lambda_B - \lambda_A = 1.00\text{ }\mu\text{m}$
$\lambda_B - \frac{\lambda_B}{3} = 1.00\text{ }\mu\text{m}$
$\frac{2}{3} \lambda_B = 1.00\text{ }\mu\text{m}$
$\lambda_B = 1.50\text{ }\mu\text{m}$