NCERT Solutions · Class 11 Physics Thermal Properties of Matter

Density is inversely tied to volume, so a positive volume-expansion coefficient produces a negative change in density — the object gets bigger, so it gets less dense.The coefficient of volume expansion \(\displaystyle \gamma \) is defined by\[\gamma = \frac{1}{V}\frac{\Delta V}{\Delta T} \]where \(\displaystyle V \) is the original volume and \(\displaystyle \Delta V \) is the change in volume for a temperature rise \(\displaystyle \Delta T \). Rearranging,\[\frac{\Delta V}{V} = \gamma \, \Delta T \]Density is mass over volume, \(\displaystyle \rho = m/V \), and the mass of the glycerine sample does not change as it is heated — only its volume does. Writing the density at the higher temperature as \(\displaystyle \rho' = m/V' \) with \(\displaystyle V' = V(1+\gamma\Delta T) \),\[\rho' = \frac{m}{V(1+\gamma\Delta T)} = \frac{\rho}{1+\gamma\Delta T} \]Since \(\displaystyle \gamma \Delta T \) works out to be small compared with $\displaystyle 1$ (check below), use the binomial approximation \(\displaystyle \dfrac{1}{1+x} \approx 1-x \):\[\rho' \approx \rho\,(1 - \gamma\Delta T) \quad\Rightarrow\quad \frac{\rho' - \rho}{\rho} = \frac{\Delta \rho}{\rho} \approx -\gamma\,\Delta T \]This is the standard result: the fractional change in density is just the negative of the fractional change in volume, because mass stays fixed while volume grows.Now substitute the given numbers. The coefficient of volume expansion for glycerine is \[\gamma = 49\times10^{-5}\ \text{K}^{-1} \] and the temperature rise is \[\Delta T = 30\ ^\circ\text{C} = 30\ \text{K} \] (a temperature interval of $\displaystyle 30$ °C equals a temperature interval of $\displaystyle 30$ K — the Celsius and Kelvin scales differ only in where zero is set, not in the size of one degree, so no conversion factor is needed here).\[\frac{\Delta \rho}{\rho} \approx -\gamma\,\Delta T = -\left(49\times10^{-5}\ \text{K}^{-1}\right)\left(30\ \text{K}\right) \]\[= -(49\times30)\times10^{-5} = -1470\times10^{-5} = -1.470\times10^{-2} \]Check that the small-quantity approximation was justified: \(\displaystyle \gamma\Delta T \approx 0.0147 \), which is indeed much less than $\displaystyle 1$, so dropping the higher-order term was valid.Both given quantities ($\displaystyle 49$ and $\displaystyle 30$) carry two significant figures, so the answer is rounded to two significant figures:\[\frac{\Delta \rho}{\rho} \approx -1.5\times10^{-2} \]The negative sign says the density decreases — glycerine expands on heating, and with the same mass spread over a larger volume, it becomes about $\displaystyle 1.5$% less dense. Fractional change is a dimensionless ratio (a change per unit of the original quantity), so it carries no unit of its own — only the sign and the magnitude matter here, not a direction in space like a vector would need.Answer: The density decreases by a fractional amount of about \(\displaystyle 1.5\times10^{-2}\) (i.e., it drops by roughly $\displaystyle 1.5$%).

The machine draws electrical power, but only half of that power actually ends up heating the aluminium — the rest is lost to the drill itself and to the surroundings. Track only the half that counts.Step $\displaystyle 1$ — Total energy the machine delivers in $\displaystyle 2.5$ minutes.Power is energy delivered per unit time, \(\displaystyle P = \dfrac{W}{t} \), so the total energy supplied is \[W = P \, t \] Convert everything to SI units first: \(\displaystyle P = 10\ \text{kW} = 1.0 \times 10^{4}\ \text{W} \), and \(\displaystyle t = 2.5\ \text{min} = 2.5 \times 60\ \text{s} = 150\ \text{s} \).\[W = (1.0 \times 10^{4}\ \text{W})(150\ \text{s}) = 1.5 \times 10^{6}\ \text{J} \]Step $\displaystyle 2$ — Only $\displaystyle 50$% of this goes into heating the block.The problem states half the power is lost to the machine and surroundings, so only half reaches the aluminium as heat: \[Q = 0.50 \times W = 0.50 \times 1.5 \times 10^{6}\ \text{J} = 7.5 \times 10^{5}\ \text{J} \]Step $\displaystyle 3$ — Relate this heat to the temperature rise using \(\displaystyle Q = mc\Delta T \).Here \(\displaystyle m \) is the mass of the block, \(\displaystyle c \) is its specific heat capacity, and \(\displaystyle \Delta T \) is the rise in temperature. This is the step where units trip people up: the specific heat is given per gram, not per kilogram, so the mass must be converted before substituting — mixing kg with a "per gram" specific heat is a factor-of-$\displaystyle 1000$ error waiting to happen.\[m = 8.0\ \text{kg} = 8.0 \times 10^{3}\ \text{g}, \qquad c = 0.91\ \text{J g}^{-1}\text{K}^{-1} \]Solving for the rise in temperature: \[\Delta T = \frac{Q}{mc} \]The denominator is the block's total heat capacity: \[mc = (8.0 \times 10^{3}\ \text{g})(0.91\ \text{J g}^{-1}\text{K}^{-1}) = 7.28 \times 10^{3}\ \text{J K}^{-1} \]Substituting: \[\Delta T = \frac{7.5 \times 10^{5}\ \text{J}}{7.28 \times 10^{3}\ \text{J K}^{-1}} = 103.0\ \text{K} \]A rise measured in kelvin is numerically identical to the same rise measured in degrees Celsius — a change in temperature doesn't shift with the offset between the two scales, only an absolute temperature would. So this is equally \(\displaystyle \Delta T = 103\ ^\circ\text{C} \).Rounding once, at the end, to three significant figures (matching the precision carried by the given data through the division):Answer: The temperature of the aluminium block rises by about $\displaystyle 103$ °C (equivalently $\displaystyle 103$ K).

Because the ice block is "large," the copper keeps losing heat until it reaches $\displaystyle 0$ °C — every joule it gives up goes into melting ice, none of it into warming meltwater, so this is the theoretical maximum.The copper block cools from $\displaystyle 500$ °C down to the ice's melting point, $\displaystyle 0$ °C. As long as there is still unmelted ice sitting in a mixture with water, the mixture's temperature is pinned at $\displaystyle 0$ °C — that is what a "large" ice block guarantees. So the copper can keep dumping heat into it until the copper itself reaches $\displaystyle 0$ °C, and that heat is used purely as latent heat, not to raise any water's temperature. That makes this the largest possible mass of ice the copper's heat can melt.Step $\displaystyle 1$ — Heat given up by the copper as it cools.Use \(\displaystyle Q = mc\Delta\theta \), where \(\displaystyle m\) is the copper's mass, \(\displaystyle c\) its specific heat capacity, and \(\displaystyle \Delta\theta\) the temperature it falls through.The specific heat is given per gram, so convert the mass to grams to match: \[m = 2.5\ \text{kg} = 2500\ \text{g} \]A rise or fall of $\displaystyle 500$ °C is the same size change as $\displaystyle 500$ K, since the Celsius and Kelvin scales have equally-sized degrees — only their zeros differ. So: \[\Delta\theta = (500 - 0)\ ^\circ\text{C} = 500\ \text{K} \]Substituting, with \(\displaystyle c = 0.39\ \text{J g}^{-1}\text{K}^{-1}\): \[Q = (2500\ \text{g})(0.39\ \text{J g}^{-1}\text{K}^{-1})(500\ \text{K}) \] \[Q = 2500 \times 0.39 \times 500\ \text{J} = 487500\ \text{J} \]Step $\displaystyle 2$ — That heat goes entirely into melting ice.The heat needed to melt a mass \(\displaystyle m_{\text{ice}}\) of ice at its melting point, with no temperature change, is \(\displaystyle Q = m_{\text{ice}} L_f \), where \(\displaystyle L_f\) is the specific latent heat of fusion. Setting the copper's heat loss equal to the ice's heat gain: \[m_{\text{ice}} = \frac{Q}{L_f} = \frac{487500\ \text{J}}{335\ \text{J g}^{-1}} \] \[m_{\text{ice}} = 1455.2\ \text{g} \]Step $\displaystyle 3$ — Round to the precision the data supports.The given mass ($\displaystyle 2.5$ kg) and specific heat ($\displaystyle 0.39$ J g⁻¹K⁻¹) each carry only $\displaystyle 2$ significant figures, so the final answer should not claim more precision than that, even though the unrounded value is $\displaystyle 1455$ g: \[m_{\text{ice}} \approx 1.5 \times 10^{3}\ \text{g} = 1.5\ \text{kg} \]This is a maximum, not a guarantee — it assumes no heat escapes to the surroundings on the way down. Any heat lost to the air around the block would melt less ice than this.Answer: The maximum mass of ice that can melt is about $\displaystyle 1.5$ kg ($\displaystyle 1.5$ × $\displaystyle 10$³ g).

When a hot object cools inside water, all the heat it gives up goes into warming the water and the calorimeter — that conservation of heat is what lets you find an unknown specific heat.Convert everything to SI units first.The metal block has mass \(\displaystyle m = 0.20 \, \text{kg} \) and cools from \(\displaystyle T_1 = 150\,^\circ\text{C} \) down to the common final temperature \(\displaystyle T = 40\,^\circ\text{C} \).The $\displaystyle 150$ cm³ of water has mass, using the density of water \(\displaystyle \rho = 1000\, \text{kg/m}^3 = 1\, \text{g/cm}^3\): \[m_w = 150\, \text{cm}^3 \times 1\, \text{g/cm}^3 = 150\, \text{g} = 0.150\, \text{kg} \]The calorimeter's water equivalent, \(\displaystyle w = 0.025\, \text{kg}\), is the step worth pausing on: it is already stated as an equivalent mass of water, so there's no need for copper's own specific heat — just add it straight onto the mass of water.Effective mass absorbing heat: \[m_w + w = 0.150\, \text{kg} + 0.025\, \text{kg} = 0.175\, \text{kg} \]This water-and-calorimeter combination starts at \(\displaystyle T_2 = 27\,^\circ\text{C}\) and is warmed to the same final temperature \(\displaystyle T = 40\,^\circ\text{C}\).Principle of calorimetry (heat lost = heat gained, assuming no heat escapes to the surroundings): the heat the metal loses while cooling equals the heat the water and calorimeter gain while warming.\[m\, c_{\text{metal}} \,(T_1 - T) = (m_w + w)\, c_w \,(T - T_2) \]Here \(\displaystyle c_{\text{metal}} \) is the specific heat capacity to be found, and \(\displaystyle c_w = 4186\, \text{J kg}^{-1}\text{K}^{-1}\) is the specific heat capacity of water (the same number whether the temperature difference is read in °C or K, since only a difference appears).Substitute the numbers. The metal cools through \(\displaystyle T_1 - T = 150 - 40 = 110\,^\circ\text{C} \) and the water warms through \(\displaystyle T - T_2 = 40 - 27 = 13\,^\circ\text{C} \):\[(0.20\, \text{kg})\, c_{\text{metal}} \,(110\, \text{K}) = (0.175\, \text{kg})(4186\, \text{J kg}^{-1}\text{K}^{-1})(13\, \text{K}) \]\[22\, c_{\text{metal}} = 9523\, \text{J K}^{-1} \]\[c_{\text{metal}} = \frac{9523}{22} = 432.9\, \text{J kg}^{-1}\text{K}^{-1} \]The masses given, \(\displaystyle 0.20\, \text{kg}\) and \(\displaystyle 0.025\, \text{kg}\), and the $\displaystyle 13$ K rise of the water, carry only two significant figures, so that is the precision the final answer can honestly claim. Rounding only now, at the end: \[c_{\text{metal}} \approx 4.3 \times 10^{2}\, \text{J kg}^{-1}\text{K}^{-1} \]For the second part: if heat leaks out to the surroundings during the experiment — through the calorimeter's walls, or while the hot block is being transferred into the water — that lost heat never shows up as a rise in the water's temperature. The calculation above assumed every joule the metal gave up went into the water and calorimeter, so it only credits the metal with the heat that actually arrived there, not the extra heat that escaped along the way. Since the metal's mass and its $\displaystyle 110$°C drop in temperature are fixed by direct measurement, and the true heat it released is larger than the $\displaystyle 9523$ J this calculation used, the true specific heat capacity of the metal must be larger than the value calculated here.Answer: The specific heat of the metal is about \(\displaystyle 4.3\times10^{2}\, \text{J kg}^{-1}\text{K}^{-1}\) ($\displaystyle 433$ J kg⁻¹ K⁻¹ before rounding). Because heat losses to the surroundings are ignored in this calculation, the value obtained is smaller than the actual specific heat of the metal.

A gas molecule's specific heat is fixed by how many independent ways it can store kinetic energy — its degrees of freedom — not by which atoms it is made of.By the law of equipartition of energy, every degree of freedom of a molecule carries an average energy of \(\displaystyle \tfrac{1}{2}k_BT \) per molecule, or \(\displaystyle \tfrac{1}{2}RT \) per mole, where \(\displaystyle R\) is the universal gas constant. So the molar internal energy is\[U = \frac{f}{2}RT \]with \(\displaystyle f\) the number of degrees of freedom. Since \(\displaystyle C_v = \left(\dfrac{\partial U}{\partial T}\right)_V\) for an ideal gas,\[C_v = \frac{f}{2}R \]Take \(\displaystyle R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}\). Converting to calories (\(\displaystyle 1\ \text{cal} = 4.184\ \text{J}\), the unit the table uses) — a step worth doing carefully, since mixing joules and calories in the same table is the easy mistake here —\[R = \frac{8.31\ \text{J mol}^{-1}\text{K}^{-1}}{4.184\ \text{J cal}^{-1}} \approx 1.99\ \text{cal mol}^{-1}\text{K}^{-1} \]Monatomic gases have \(\displaystyle f = 3\). A single atom is a point mass: it has only the three translational degrees of freedom (motion along \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\)). There is no bond to rotate about that stores energy, and nothing to vibrate. So\[C_v = \frac{3}{2}R = \frac{3}{2}(1.99\ \text{cal mol}^{-1}\text{K}^{-1}) \approx 2.98\ \text{cal mol}^{-1}\text{K}^{-1} \]which is essentially the \(\displaystyle 2.92\ \text{cal mol}^{-1}\text{K}^{-1}\) quoted for monatomic gases — the small gap between $\displaystyle 2.98$ and $\displaystyle 2.92$ is the usual real-gas correction to the ideal-gas equipartition value.Diatomic gases have \(\displaystyle f = 5\). A diatomic molecule such as \(\displaystyle \mathrm{H_2}\), \(\displaystyle \mathrm{N_2}\), \(\displaystyle \mathrm{O_2}\), NO or CO is a rigid dumbbell. Besides its $\displaystyle 3$ translational degrees of freedom, it can rotate about the two axes perpendicular to the bond axis; rotation about the bond axis itself is not counted, because the moment of inertia of two point-like atoms about their own line is essentially zero and stores no energy. That adds $\displaystyle 2$ rotational degrees of freedom, so \(\displaystyle f = 5\):\[C_v = \frac{5}{2}R = \frac{5}{2}(1.99\ \text{cal mol}^{-1}\text{K}^{-1}) \approx 4.97\ \text{cal mol}^{-1}\text{K}^{-1} \]This is exactly the band the table shows for \(\displaystyle \mathrm{H_2}\), \(\displaystyle \mathrm{N_2}\), \(\displaystyle \mathrm{O_2}\), NO and CO — \(\displaystyle 4.87\) to \(\displaystyle 5.02\ \text{cal mol}^{-1}\text{K}^{-1}\). It tells us that at room temperature these molecules are translating and rotating freely, but their bond is not yet vibrating: the vibrational degree of freedom is "frozen out," because the spacing between vibrational energy levels is much larger than \(\displaystyle k_BT\) at $\displaystyle 300$ K, so the equipartition theorem does not yet apply to that motion.Chlorine's larger value means its bond is already vibrating at room temperature. A vibrating bond adds two more degrees of freedom — one kinetic (motion along the bond) and one potential (energy stored in the stretched or compressed bond), each contributing \(\displaystyle \tfrac{1}{2}R\) — giving a fully-active vibrational diatomic \(\displaystyle f = 7\):\[C_v = \frac{7}{2}R = \frac{7}{2}(1.99\ \text{cal mol}^{-1}\text{K}^{-1}) \approx 6.96\ \text{cal mol}^{-1}\text{K}^{-1} \quad \text{(vibration fully classical)} \]Chlorine's measured value, \(\displaystyle 6.17\ \text{cal mol}^{-1}\text{K}^{-1}\), falls strictly between the rigid-rotator value \(\displaystyle \tfrac{5}{2}R \approx 4.97\ \text{cal mol}^{-1}\text{K}^{-1}\) and this fully-vibrating value \(\displaystyle \tfrac{7}{2}R \approx 6.96\ \text{cal mol}^{-1}\text{K}^{-1}\). Reading the effective number of degrees of freedom back out of the measured heat capacity,\[f = \frac{2C_v}{R} = \frac{2(6.17\ \text{cal mol}^{-1}\text{K}^{-1})}{1.99\ \text{cal mol}^{-1}\text{K}^{-1}} \approx 6.2 \]so chlorine behaves as though it has about $\displaystyle 6.2$ degrees of freedom, not a clean $\displaystyle 5$ or $\displaystyle 7$ — its vibrational mode is partially excited, neither silent nor fully classical.The physical reason is mass and bond stiffness. Chlorine atoms are far heavier than hydrogen, nitrogen, oxygen or carbon atoms, and the \(\displaystyle \mathrm{Cl}\text{-}\mathrm{Cl}\) bond is comparatively weak, so the molecule's natural vibration frequency \(\displaystyle \nu\) is low. A low \(\displaystyle \nu\) means a small vibrational energy quantum \(\displaystyle h\nu\), one that is already comparable to \(\displaystyle k_BT\) at room temperature — so a significant fraction of chlorine molecules occupy excited vibrational states, and the vibrational mode contributes to \(\displaystyle C_v\) even though it has not yet reached the fully classical limit. The lighter, more tightly bound molecules \(\displaystyle \mathrm{H_2}\), \(\displaystyle \mathrm{N_2}\), \(\displaystyle \mathrm{O_2}\), NO and CO have much higher vibrational frequencies, so \(\displaystyle h\nu \gg k_BT\) at $\displaystyle 300$ K, essentially no molecules reach the first vibrational level, and that mode stays frozen out of their heat capacity.Answer: Monatomic gases have only $\displaystyle 3$ translational degrees of freedom, giving \(\displaystyle C_v = \tfrac{3}{2}R \approx 2.98\ \text{cal mol}^{-1}\text{K}^{-1}\), close to the observed 2.92. Diatomic gases add $\displaystyle 2$ rotational degrees of freedom (\(\displaystyle f = 5\)), giving \(\displaystyle C_v = \tfrac{5}{2}R \approx 4.97\ \text{cal mol}^{-1}\text{K}^{-1}\), matching \(\displaystyle \mathrm{H_2}\), \(\displaystyle \mathrm{N_2}\), \(\displaystyle \mathrm{O_2}\), NO and CO, whose vibrational mode is frozen out at room temperature. Chlorine's higher value, \(\displaystyle 6.17\ \text{cal mol}^{-1}\text{K}^{-1}\) (between \(\displaystyle \tfrac{5}{2}R\) and the fully-vibrating \(\displaystyle \tfrac{7}{2}R \approx 6.96\ \text{cal mol}^{-1}\text{K}^{-1}\)), shows that its vibrational degree of freedom is already partially excited at room temperature — because its heavier atoms and weaker bond give it a low vibrational frequency, unlike the lighter diatomic gases.

A change of $\displaystyle 3$°F is not a change of $\displaystyle 3$°C — you must scale the temperature drop before using it in \(\displaystyle Q = mc\Delta T\).The Fahrenheit and Celsius scales have different degree sizes, so a difference in temperature converts with the same factor of \(\displaystyle 5/9\) that the full conversion formula uses:\[\Delta T(^\circ\text{C}) = \frac{5}{9}\,\Delta T(^\circ\text{F}) \]Here the fever drops from $\displaystyle 101$°F to $\displaystyle 98$°F, so\[\Delta T(^\circ\text{F}) = 101 - 98 = 3\ ^\circ\text{F} \]\[\Delta T(^\circ\text{C}) = \frac{5}{9}\times 3 = \frac{5}{3}\ ^\circ\text{C} \approx 1.667\ ^\circ\text{C} \]Step $\displaystyle 1$: Heat that must leave the body for this temperature drop.The body's specific heat capacity is taken equal to that of water, \(\displaystyle c = 1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}\). Using\[Q = mc\,\Delta T \]where \(\displaystyle m\) is the mass of the child (in grams) and \(\displaystyle c\) the specific heat, substitute \(\displaystyle m = 30\ \text{kg} = 3\times10^{4}\ \text{g}\):\[Q = (3\times10^{4}\ \text{g})(1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1})\left(\frac{5}{3}\ ^\circ\text{C}\right) \]\[Q = 5\times10^{4}\ \text{cal} = 50000\ \text{cal} \]Step $\displaystyle 2$: Convert this heat into a mass of sweat evaporated.The problem tells you to treat evaporation as the only mechanism by which this heat is lost, so all of \(\displaystyle Q\) goes into vaporising sweat. Using the latent heat relation\[Q = mL \]where \(\displaystyle L = 580\ \text{cal g}^{-1}\) is the latent heat of evaporation of water at this temperature, solve for the mass \(\displaystyle m\) of water evaporated:\[m = \frac{Q}{L} = \frac{50000\ \text{cal}}{580\ \text{cal g}^{-1}} = 86.2\ \text{g} \]Step $\displaystyle 3$: Turn this into a rate.This mass evaporates over the $\displaystyle 20$ minutes it takes the fever to fall, so the average extra rate of evaporation is\[R = \frac{m}{t} = \frac{86.2\ \text{g}}{20\ \text{min}} \approx 4.31\ \text{g min}^{-1} \]Equivalently, since $\displaystyle 20$ min = $\displaystyle 1200$ s,\[R = \frac{86.2\ \text{g}}{1200\ \text{s}} \approx 0.0718\ \text{g s}^{-1} \]The input data ($\displaystyle 30$ kg, $\displaystyle 20$ min, $\displaystyle 580$ cal g⁻¹) carry two to three significant figures, so the result is rounded once, at the end, to three significant figures: \(\displaystyle 4.31\ \text{g min}^{-1}\).Answer: The drug causes an average extra evaporation rate of about \(\displaystyle 4.31\ \text{g min}^{-1}\) (≈ $\displaystyle 0.0718$ g s⁻¹), corresponding to roughly $\displaystyle 86.2$ g of sweat evaporated over the $\displaystyle 20$ minutes.

The rate of heat conduction depends on the total surface area exposed, and a cube has six faces all leaking heat inward at once — not one.Heat flows into the box by conduction, following \[H = \frac{Q}{t} = \frac{kA\,\Delta T}{x} \] where \(\displaystyle k\) is the thermal conductivity of the wall material, \(\displaystyle A\) is the total area through which heat flows, \(\displaystyle \Delta T\) is the temperature difference across the wall, \(\displaystyle x\) is the wall thickness, and \(\displaystyle t\) is the time.Step $\displaystyle 1$ — Total conducting area. The box is a cube of side \(\displaystyle L = 30\ \text{cm} = 0.30\ \text{m}\). Heat enters through all six faces simultaneously, so \[A = 6L^{2} = 6\times(0.30\ \text{m})^{2} = 6\times 0.09\ \text{m}^{2} = 0.54\ \text{m}^{2} \] Using the area of just one face here is the mistake to avoid — it would understate the heat gained by a factor of six.Step $\displaystyle 2$ — Temperature difference and thickness. Outside temperature is \(\displaystyle 45\,^{\circ}\text{C}\). As long as ice is still present inside, the inner wall surface stays pinned at the ice–water mixture's melting point, \(\displaystyle 0\,^{\circ}\text{C}\) — it does not drift upward while ice and liquid water coexist. So \[\Delta T = 45\,^{\circ}\text{C} - 0\,^{\circ}\text{C} = 45\ \text{K} \] (a difference of so many degrees Celsius is numerically the same difference in kelvin). The wall thickness is \(\displaystyle x = 5.0\ \text{cm} = 0.05\ \text{m}\).Step $\displaystyle 3$ — Time in SI units. The conductivity is given per second, so the $\displaystyle 6$ h must be converted: \[t = 6\ \text{h} \times 3600\ \text{s h}^{-1} = 21600\ \text{s} \]Step $\displaystyle 4$ — Total heat conducted in. \[Q = \frac{kA\,\Delta T\,t}{x} = \frac{(0.01\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1})(0.54\ \text{m}^{2})(45\ \text{K})(21600\ \text{s})}{0.05\ \text{m}} \] Multiplying the numerator through: \[0.01 \times 0.54 = 0.0054,\qquad 0.0054 \times 45 = 0.243,\qquad 0.243 \times 21600 = 5248.8 \] \[Q = \frac{5248.8\ \text{J}}{0.05} = 104976\ \text{J} \]Step $\displaystyle 5$ — Turn that heat into a mass of melted ice. This heat does not raise any temperature — it goes entirely into changing ice at \(\displaystyle 0\,^{\circ}\text{C}\) into water at \(\displaystyle 0\,^{\circ}\text{C}\), which is exactly what latent heat describes: \[Q = mL_f \quad\Rightarrow\quad m = \frac{Q}{L_f} \] with \(\displaystyle L_f = 335\times10^{3}\ \text{J kg}^{-1}\), the heat of fusion of water. So \[m = \frac{104976\ \text{J}}{335\times10^{3}\ \text{J kg}^{-1}} = 0.3134\ \text{kg} \]Step $\displaystyle 6$ — Subtract from the original ice. \[\text{ice remaining} = 4.0\ \text{kg} - 0.3134\ \text{kg} = 3.6866\ \text{kg} \] The given data ($\displaystyle 4.0$ kg, $\displaystyle 30$ cm, $\displaystyle 5.0$ cm, $\displaystyle 45$ °C) are all stated to two significant figures, so the result is rounded to two significant figures, not carried at full calculator precision: \[\text{ice remaining} \approx 3.7\ \text{kg} \]Answer: About $\displaystyle 3.7$ kg of ice is left after $\displaystyle 6$ h (roughly $\displaystyle 0.31$ kg melts in that time).

Steady heat conduction through the brass base must carry heat into the water at exactly the rate the water is boiling away — setting the "heat conducted per second" equal to the "heat needed to vaporise per second" pins down the flame-side temperature.Two laws are needed.Fourier's law of heat conduction: for a slab of thickness \(\displaystyle d\) and cross-sectional area \(\displaystyle A\), with its two faces at temperatures \(\displaystyle T_1\) (hotter) and \(\displaystyle T_2\) (cooler), the rate of heat flow through it is \[\frac{dQ}{dt} = \frac{K A (T_1 - T_2)}{d} \] where \(\displaystyle K\) is the thermal conductivity of the material. Here \(\displaystyle T_1\) is the temperature of the flame-side face of the boiler's base, and \(\displaystyle T_2\) is the temperature of the water-side face.Latent heat of vaporisation: converting a mass \(\displaystyle m\) of liquid to vapour, at the boiling point, takes heat \(\displaystyle Q = mL\), where \(\displaystyle L\) is the latent heat of vaporisation. So the rate of heat delivery needed to keep boiling water away at a steady rate is \[\frac{dQ}{dt} = \frac{dm}{dt}\,L \]This is the step people slip on: boiling happens at constant temperature, so none of this heat raises the water's temperature — it all goes into the liquid-to-vapour change of state. That is exactly why \(\displaystyle L\), not a specific heat capacity, belongs in this equation.Putting in the numbersThe base of the boiler is the conducting slab: \(\displaystyle A = 0.15\ m^2\), \(\displaystyle d = 1.0\ cm = 1.0\times10^{-2}\ m\), \(\displaystyle K = 109\ J\,s^{-1}m^{-1}K^{-1}\).Water is boiled off at \[\frac{dm}{dt} = 6.0\ kg/min = \frac{6.0}{60}\ kg/s = 0.10\ kg/s \] with \(\displaystyle L = 2256\times10^{3}\ J\,kg^{-1}\).So the heat current that must be crossing the base is \[\frac{dQ}{dt} = (0.10\ kg/s)(2256\times10^{3}\ J/kg) = 2.256\times10^{5}\ J/s \]Solving for the flame-side temperatureThe inner face of the base sits in contact with boiling water, so \(\displaystyle T_2 = 100\,^\circ\!C\) — water boils at \(\displaystyle 100\,^\circ\!C\) at atmospheric pressure, and as long as it is steadily boiling, that face stays pinned at that temperature.Equating the two expressions for \(\displaystyle dQ/dt\): \[\frac{K A (T_1 - T_2)}{d} = 2.256\times10^{5}\ J/s \] \[T_1 - T_2 = \frac{(2.256\times10^{5}\ J/s)(1.0\times10^{-2}\ m)}{(109\ J\,s^{-1}m^{-1}K^{-1})(0.15\ m^2)} = \frac{2.256\times10^{3}}{16.35}\ K \approx 138\ K \]A temperature difference of $\displaystyle 138$ kelvin is the same size as a difference of $\displaystyle 138$ degrees Celsius — the two scales differ only in where zero is set, not in the size of one degree — so this number can be added straight onto a Celsius temperature.\[T_1 = 100\,^\circ\!C + 138\,^\circ\!C \approx 238\,^\circ\!C \]The given quantities carry two or three significant figures ($\displaystyle 0.15$, $\displaystyle 1.0$, $\displaystyle 6.0$ to two; $\displaystyle 109$ to three), so the result is stated to three figures. Note also that this is only an estimate of the flame temperature right at the contact point: not all the heat leaving the flame enters the boiler — some escapes past its sides — so the true flame temperature is at least this high.Answer: The part of the flame in contact with the boiler is at approximately \(\displaystyle 238\,^\circ\!C\) (about $\displaystyle 138$ °C above the boiling point of water).

Emission and absorption are two sides of the same coin — a surface that reflects instead of absorbing radiation cannot radiate it back either.(a) A body with large reflectivity is a poor emitterKirchhoff's law of radiation says that at a given temperature, the ratio of a surface's emissive power to its absorptive power equals the emissive power of an ideal black body at that same temperature: \[\frac{e(\lambda,T)}{a(\lambda,T)} = E_{\text{black}}(\lambda,T) \] so \(\displaystyle e = a\,E_{\text{black}} \): emissivity and absorptivity always track together. A highly reflective surface, by definition, sends most of the radiation falling on it back out unabsorbed — its absorptivity \(\displaystyle a\) is small. Since \(\displaystyle e\) is proportional to \(\displaystyle a\), a surface with low absorptivity must also have low emissivity. That is why polished, silvery surfaces (used on thermos flasks, for instance) both reflect radiation well and radiate heat poorly, while dull black surfaces absorb well and emit well.(b) A brass tumbler feels much colder than a wooden tray on a chilly dayOn a chilly day, both objects sit in the same room and are at the same temperature — below your skin's temperature. What differs is thermal conductivity. Brass is a good conductor, so the instant your hand touches it, heat from your skin is conducted away into the bulk of the metal very rapidly, and your nerve endings register a fast, large temperature drop — this is what "feels cold." Wood is a poor conductor (an insulator); it carries heat away from your hand only very slowly, so the touch does not feel cold even though the wood is at the identical ambient temperature. The sensation of "hot" or "cold" on touch is really a measure of the rate of heat flow out of your skin, not the object's actual temperature — this is the trap in the question.(c) A black-body-calibrated pyrometer under-reads a red-hot iron piece in the open, but reads it correctly inside a furnaceAn optical pyrometer works by matching the visible-radiation intensity from the hot object to that of an ideal black body, using the Stefan–Boltzmann relation for an ideal radiator, \(\displaystyle E = \sigma T^4 \). A real iron piece is not an ideal black body: its emissivity \(\displaystyle e<1\), so it actually radiates only \(\displaystyle e\sigma T^4 \), less than a black body at the same true temperature \(\displaystyle T\) would. The pyrometer, calibrated on the assumption \(\displaystyle e=1\), sees this weaker radiation and infers a lower temperature than the iron's real temperature — hence "too low" a reading in the open.Inside a furnace, the situation changes because the furnace cavity itself acts as a near-ideal black body. Radiation inside an enclosed cavity at uniform temperature bounces off the walls many times, being partially absorbed at each reflection, until essentially all of it is absorbed and re-emitted — this multiple-reflection trapping makes the radiation coming out of a small opening in the furnace wall closely match ideal black-body radiation at the furnace temperature, regardless of the iron's own surface emissivity. So inside the furnace the pyrometer is effectively looking at black-body radiation, matching its calibration, and it reads the correct temperature.(d) The earth without its atmosphere would be inhospitably coldThe earth's surface, warmed by sunlight during the day, radiates energy back out mainly as infrared radiation. The atmosphere — chiefly its carbon dioxide and water vapour — absorbs a large part of this outgoing infrared radiation and re-radiates part of it back down to the surface (the greenhouse effect), so heat is retained near the surface through the night. Without an atmosphere, none of this outgoing radiation would be trapped: the surface would lose heat freely to space as soon as sunlight disappeared, and night-time temperatures would plunge drastically, much as they do on the Moon (which has no atmosphere and swings between roughly \(\displaystyle +400\text{ K}\) in sunlight to about \(\displaystyle 100\text{ K}\) in shadow). The atmosphere is what keeps the earth's surface temperature in the habitable range rather than oscillating between extremes.(e) Steam heating systems warm a building more efficiently than hot-water heatingCirculating hot water delivers heat to a room only by cooling down: the heat given up per unit mass is \(\displaystyle Q = mc\,\Delta T \), where \(\displaystyle c\) is water's specific heat capacity and \(\displaystyle \Delta T\) is the drop in temperature of the water as it passes through the radiator. Steam, by contrast, arrives at \(\displaystyle 100\ ^\circ\text{C}\) and gives up a very large additional quantity of heat as it condenses back into water at the same temperature — the latent heat of vaporization, \(\displaystyle L \approx 22.6 \times 10^{5}\ \text{J kg}^{-1} \), released via \(\displaystyle Q = mL \). This latent-heat contribution is far larger than the sensible heat a similar mass of hot water can give up for a modest temperature drop. So for the same mass of fluid circulated, steam delivers substantially more heat to the room, making steam-based heating systems more efficient than hot-water ones.Answer: (a) low absorptivity ⇒ low emissivity by Kirchhoff's law; (b) brass conducts heat away from the hand faster than wood, though both are at the same temperature; (c) the iron's emissivity is less than $\displaystyle 1$ in the open (pyrometer under-reads) but the furnace cavity itself radiates like an ideal black body (correct reading); (d) without the atmosphere's greenhouse trapping of outgoing infrared radiation, the surface would radiate all its heat to space and become extremely cold; (e) condensing steam releases its large latent heat of vaporization in addition to cooling, delivering far more heat than hot water cooling by the same temperature drop.

Newton's law of cooling, used over a finite time interval, compares the average temperature of the body to the temperature of the surroundings — not the instantaneous temperature at either end.For a body cooling slowly enough that Newton's law applies, the useful working form is \[\frac{\theta_1-\theta_2}{t}=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) \] where \(\displaystyle \theta_1\) is the initial temperature, \(\displaystyle \theta_2\) the final temperature, \(\displaystyle t\) the time taken to cool between them, \(\displaystyle \theta_0\) the (constant) temperature of the surroundings, and \(\displaystyle K\) a constant fixed by the body's surface and the surrounding conditions — not by the temperatures themselves. Because \(\displaystyle K\) does not change between the two coolings described here (same body, same surroundings), it can be found from the first cooling and then reused for the second.Step $\displaystyle 1$: find \(\displaystyle K\) from the first cooling ($\displaystyle 80$ °C → $\displaystyle 50$ °C in $\displaystyle 5$ min).Here \(\displaystyle \theta_1=80\ ^\circ\text{C}\), \(\displaystyle \theta_2=50\ ^\circ\text{C}\), \(\displaystyle t=5\ \text{min}\), \(\displaystyle \theta_0=20\ ^\circ\text{C}\). Substituting, \[\frac{80-50}{5}=K\left(\frac{80+50}{2}-20\right) \] \[\frac{30}{5}=K(65-20) \] \[6=K(45) \] \[K=\frac{6}{45}=\frac{2}{15}\ \text{min}^{-1} \]A common slip here is to use the instantaneous form \(\displaystyle d\theta/dt=-K(\theta-\theta_0)\) with just the starting temperature; because the body's temperature is changing throughout the $\displaystyle 5$ minutes, the law is applied to the average temperature over the interval, \(\displaystyle (\theta_1+\theta_2)/2\), not to \(\displaystyle \theta_1\) alone.Step $\displaystyle 2$: use the same \(\displaystyle K\) for the second cooling ($\displaystyle 60$ °C → $\displaystyle 30$ °C).Now \(\displaystyle \theta_1=60\ ^\circ\text{C}\), \(\displaystyle \theta_2=30\ ^\circ\text{C}\), \(\displaystyle \theta_0=20\ ^\circ\text{C}\) (the surroundings have not changed), and \(\displaystyle t\) is unknown: \[\frac{60-30}{t}=K\left(\frac{60+30}{2}-20\right) \] \[\frac{30}{t}=K(45-20) \] \[\frac{30}{t}=25K \] \[t=\frac{30}{25K}=\frac{30}{25\times\dfrac{2}{15}}=\frac{30}{\dfrac{50}{15}}=\frac{30\times15}{50} \] \[t=\frac{450}{50}=9\ \text{min} \]All the given data ($\displaystyle 80$, $\displaystyle 50$, $\displaystyle 60$, $\displaystyle 30$, $\displaystyle 20$ °C and $\displaystyle 5$ min) carry two significant figures, so the answer is stated to two significant figures: $\displaystyle 9.0$ minutes. No rounding was needed along the way since the numbers divide out exactly.It is worth noting why the second cooling takes longer even though it spans the same $\displaystyle 30$ °C drop as the first: the body is closer to the surroundings' temperature (its average excess is \(\displaystyle 45-20=25\ ^\circ\text{C}\) instead of \(\displaystyle 65-20=45\ ^\circ\text{C}\)), so it cools more slowly, exactly as Newton's law predicts.Answer: $\displaystyle 9.0$ minutes.