Basic of Heat Transfer and Conduction Questions in English

(D) In a steady state,the temperature at any given point within a body does not change with time.
This occurs because the rate of heat entering a specific region of the body is exactly equal to the rate of heat leaving that region.
However,the temperature can vary from one point to another within the body,depending on the thermal gradient and the distance from the heat source.
Therefore,the correct statement is that the temperature does not change with time but is different at different points of the body.

(D) When a rod reaches a steady state,the temperature at every point along the rod remains constant over time,meaning $\frac{dT}{dt} = 0$ for all points.
This does not mean heat flow has stopped; rather,it means that for any small cross-section of the rod,the rate of heat entering the section is exactly equal to the rate of heat leaving the section.
Therefore,each point of the rod is transferring heat to its neighbor at the same rate at which it is receiving heat from the previous point.
Thus,option $D$ is correct.

(A) When a piece of glass is heated or cooled rapidly,the surface temperature changes much faster than the interior temperature due to its low thermal conductivity.
This creates a temperature gradient within the material,leading to unequal expansion or contraction of its layers.
Since glass is a brittle material,these internal stresses exceed its structural limit,causing the glass to crack.

(B) The temperature gradient is defined as the change in temperature per unit length,given by the formula: $\text{Temperature Gradient} = \frac{\theta_1 - \theta_2}{l}$.
Here,$\theta_1 = 30 \ ^oC$ is the temperature of the hotter end,$l = 0.5 \ m$ is the length of the rod,and the temperature gradient is $80 \ ^oC/m$.
Substituting the values into the formula:
$80 = \frac{30 - \theta_2}{0.5}$.
Multiplying both sides by $0.5$:
$80 \times 0.5 = 30 - \theta_2$.
$40 = 30 - \theta_2$.
Rearranging to solve for $\theta_2$:
$\theta_2 = 30 - 40 = -10 \ ^oC$.
Therefore,the temperature of the cooler end is $-10 \ ^oC$.

(A) Snow consists of fine ice crystals with a large amount of air trapped between them. Since air is a very poor conductor of heat (an insulator),the presence of trapped air makes snow a much better heat insulator than solid ice,which lacks these air pockets. Therefore,the correct reason is that air is trapped in the pores of snow.

(B) Two thin blankets are usually warmer than one thick blanket because air is trapped between them.
Air is a bad conductor of heat.
This trapped layer of air prevents the transfer of heat from our body to the surroundings,thereby keeping us warmer.

(B) According to Fourier's law of heat conduction,the rate of heat flow $\frac{dQ}{dt}$ through a material is given by the formula: $\frac{dQ}{dt} = -KA \frac{d\theta}{dx}$.
Here,$K$ is the thermal conductivity,$A$ is the area,and $\frac{d\theta}{dx}$ is the temperature gradient.
For a unit area $(A = 1)$,the quantity of heat crossing the area per unit time is proportional to the temperature gradient $\frac{d\theta}{dx}$ perpendicular to the surface.
Therefore,the correct option is $B$.

(B) In a thermal steady state,the temperature gradient along a uniform rod is constant.
The temperature gradient is given by: $\frac{dT}{dx} = \frac{T_{hot} - T_{cold}}{L} = \frac{100^\circ C - 20^\circ C}{20 \ cm} = \frac{80}{20} = 4^\circ C/cm$.
The temperature at any distance $x$ from the hot end is given by $T(x) = T_{hot} - (\text{gradient} \times x)$.
At the centre of the rod,$x = 10 \ cm$.
Therefore,$T(10) = 100^\circ C - (4^\circ C/cm \times 10 \ cm) = 100^\circ C - 40^\circ C = 60^\circ C$.

(C) Heat always flows from a region of higher temperature to a region of lower temperature.
Therefore,for heat to flow from one end of a solid to the other,there must be a difference in temperature along its length,which is known as a temperature gradient.

(A) The temperature gradient is defined as the rate of change of temperature with respect to distance along the rod.
It is given by the formula: $\text{Temperature gradient} = \frac{\Delta T}{\Delta x}$.
Here,the temperature difference $\Delta T = 125^\circ C - 25^\circ C = 100^\circ C$.
The length of the rod $\Delta x = 50 \ cm$.
Therefore,the temperature gradient $= \frac{100^\circ C}{50 \ cm} = 2^\circ C/cm$.

(C) The correct option is $C$.
In winter,the ambient temperature is significantly lower than the human body temperature (approximately $37.4 ^\circ C$).
Woolen clothes contain a large amount of trapped air between their fibers.
Since air is a very poor conductor of heat and wool itself is a bad conductor,these clothes act as an insulator.
This prevents the heat generated by the human body from escaping into the colder surroundings,thereby keeping the body warm.

(A) Convection occurs due to the difference in density caused by gravity,where warmer,less dense fluid rises and cooler,denser fluid sinks.
In a state of weightlessness,there is no effective gravity to create these buoyancy-driven currents.
Therefore,convection is not possible.
Heat transfer in the liquid will occur primarily through conduction.

(D) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K \theta}{\rho L x}$,where $K$ is thermal conductivity,$\rho$ is density,$L$ is latent heat,and $\theta$ is the temperature difference.
Integrating this,the time $t$ taken to grow from thickness $x_1$ to $x_2$ is $t = \frac{\rho L}{2K\theta} (x_2^2 - x_1^2)$.
This implies $t \propto (x_2^2 - x_1^2)$.
For the first $1 \; cm$ ($0$ to $1 \; cm$): $7 \propto (1^2 - 0^2) = 1$.
For the next $1 \; cm$ ($1 \; cm$ to $2 \; cm$): $t' \propto (2^2 - 1^2) = 4 - 1 = 3$.
Comparing the two,$\frac{t'}{7} = \frac{3}{1}$,which gives $t' = 21$ hours.
Since $21 > 7$,the time taken is more than $7$ hours.

(C) The rate of heat flow through a conductor is given by Fourier's law of heat conduction:
$\frac{dQ}{dt} = -KA \frac{dT}{dx}$
Since the heat flow rate $\frac{dQ}{dt}$,thermal conductivity $K$,and cross-sectional area $A$ are constants for a uniform conductor,we have:
$\frac{dT}{dx} = -\frac{1}{KA} \frac{dQ}{dt} = \text{constant}$
This implies that the temperature gradient $\frac{dT}{dx}$ is constant.
Integrating this expression,we get $T(x) = -mx + c$,where $m$ is a positive constant.
This represents a linear decrease in temperature with respect to distance $x$. Therefore,the graph of $T$ versus $x$ is a straight line with a negative slope,which corresponds to Graph $C$.

(B) In a steady state,the rate of heat flow $(H)$ through any cross-section of a conductor is constant,provided there is no heat loss from the curved surface. Since the problem states that the curved surface is thermally isolated from the surroundings,no heat escapes through the sides. Therefore,the heat flow rate $H$ remains the same at every cross-section along the length of the conductor. Consequently,the graph of $H$ versus $x$ will be a straight line parallel to the $x$-axis.

(D) When a rod reaches a steady state,the temperature at every point along the rod remains constant over time,meaning $\frac{dT}{dt} = 0$. In this state,the amount of heat entering any cross-section of the rod per unit time must equal the amount of heat leaving it. Therefore,every point conducts the same amount of heat to its neighboring point at a constant rate.

(D) In the steady state,the temperature at any given point within a body does not change with time because the rate of heat flow into that point is equal to the rate of heat flow out of that point.
However,the temperature can vary from one point to another within the body depending on the thermal gradient and the boundary conditions.
Therefore,the temperature is independent of time but dependent on the position within the body.

(D) Under steady-state conditions,the rate of heat flow $H$ through a rod is constant and is given by the formula:
$H = \frac{KA(T_1 - T_2)}{L} = \frac{KA(T_1 - T_x)}{x}$
where $T_1$ is the temperature at the hot end $(x=0)$,$T_2$ is the temperature at the cold end $(x=L)$,and $T_x$ is the temperature at distance $x$.
Rearranging the equation for $T_x$:
$T_x = T_1 - \left( \frac{T_1 - T_2}{L} \right) x$
This is a linear equation of the form $y = mx + c$,where the slope $m = -\frac{T_1 - T_2}{L}$ is negative and constant.
Therefore,the temperature $\theta$ decreases linearly with distance $x$,which corresponds to a straight line with a negative slope.

(A) The temperature gradient is defined as the change in temperature per unit length of the rod.
It is given by the formula: $\text{Temperature Gradient} = \frac{\Delta \theta}{\Delta x}$
Where $\Delta \theta = \theta_2 - \theta_1 = 125 \ ^\circ C - 25 \ ^\circ C = 100 \ ^\circ C$.
And the length of the rod $\Delta x = 50 \ cm$.
Therefore,$\text{Temperature Gradient} = \frac{100 \ ^\circ C}{50 \ cm} = 2 \ ^\circ C/cm$.

(A) The temperature gradient is defined as the rate of change of temperature with respect to distance.
Temperature gradient $= \frac{\Delta \theta}{\Delta x} = \frac{\theta_2 - \theta_1}{\Delta x}$
Given: $\Delta \theta = 125^oC - 25^oC = 100^oC$ and $\Delta x = 50 \, cm$.
Temperature gradient $= \frac{100}{50} = 2 \, ^oC/cm$.

(D) The time $t$ required for the thickness of an ice layer to increase from $y_1$ to $y_2$ is given by the formula: $t = \frac{\rho L}{2K\theta} (y_2^2 - y_1^2)$.
For the first layer ($0 \, cm$ to $1 \, cm$):
$t_1 = C(1^2 - 0^2) = C(1) = 7 \, h$,where $C = \frac{\rho L}{2K\theta}$.
For the second layer ($1 \, cm$ to $2 \, cm$):
$t_2 = C(2^2 - 1^2) = C(4 - 1) = 3C$.
Since $C = 7 \, h$,we have:
$t_2 = 3 \times 7 = 21 \, h$.

(A) In a steady state of thermal conduction,the temperature gradient along the rod is constant.
Let $\theta_A = 100^\circ C$ be the temperature at end $A$ and $\theta_B = 0^\circ C$ be the temperature at end $B$.
The total length of the rod is $L = 20 \, cm$.
Let $\theta_x$ be the temperature at a point $x = 6 \, cm$ from end $A$.
The temperature gradient is given by $\frac{\theta_A - \theta_x}{x} = \frac{\theta_A - \theta_B}{L}$.
Substituting the given values:
$\frac{100 - \theta_x}{6} = \frac{100 - 0}{20}$
$\frac{100 - \theta_x}{6} = \frac{100}{20}$
$100 - \theta_x = 5 \times 6$
$100 - \theta_x = 30$
$\theta_x = 100 - 30 = 70^\circ C$.
Thus,the temperature at the point is $70^\circ C$.

(D) The rate of heat transfer through a rod depends on its thermal conductivity $(K)$.
Silver,copper,and brass are metals with high thermal conductivity,meaning they conduct heat away from the paper rapidly,preventing the paper from reaching its ignition temperature quickly.
Wood is an insulator with very low thermal conductivity.
Because wood does not conduct heat away from the surface,the heat supplied to the paper remains localized at the point of contact.
Therefore,the paper wrapped around the wooden rod will reach its ignition temperature first and burn.

(A) thermocouple is a device consisting of two dissimilar electrical conductors forming an electrical junction. $A$ thermocouple produces a temperature-dependent voltage as a result of the Seebeck effect,which is a form of converting heat energy into electrical energy. This voltage can be interpreted to measure temperature.

(A) In the steady state,the rate of heat flow $Q$ through a rod of cross-sectional area $A$ and thermal conductivity $K$ is given by Fourier's law of heat conduction:
$Q = -KA \frac{dT}{dx}$
Since the rod is uniform and in a steady state,$Q$,$K$,and $A$ are constants.
Therefore,$\frac{dT}{dx} = -\frac{Q}{KA} = \text{constant}$.
This implies that the temperature gradient is constant throughout the rod.
$A$ constant negative slope indicates a straight line starting from $T = 100\,^{\circ}C$ at $x = 0$ and decreasing linearly to $T = 0\,^{\circ}C$ at the other end $B$.

(A) Woolen clothes have a porous structure that traps air between the fibers.
Since air is a bad conductor of heat,it prevents the heat from the human body from escaping into the cold surroundings.
Therefore,the trapped air acts as an insulator,keeping the body warm in winter.
Both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(A) When two thin blankets are placed together,a layer of air is trapped between them.
Since air is a very poor conductor of heat (a good insulator),this trapped air layer prevents the heat from the body from escaping to the surroundings.
Therefore,the combination of two thin blankets provides better insulation and warmth compared to a single thick blanket of the same total thickness,as the air layer acts as an additional thermal barrier.

(N/A) Heat is a form of energy.
Heat transfer is the process of energy transfer from one system to another,or from one part of a system to another,arising due to a temperature difference.
Thus,the movement of heat within a substance or between substances is called heat transfer.
There are $3$ distinct modes of heat transfer:
$(1)$ Conduction: Heat transfer in solids,where energy is transferred through molecular collisions without the actual movement of matter.
$(2)$ Convection: Heat transfer in fluids (liquids and gases),which involves the actual movement of the heated particles of the fluid.
$(3)$ Radiation: Heat transfer that does not require any material medium; it occurs through electromagnetic waves.

(N/A) Thermal Conduction: Conduction is the mechanism of heat transfer between two adjacent parts of a body due to a temperature difference between them.
In solid bodies,thermal conduction occurs primarily through two mechanisms: the vibration of atoms or molecules about their mean positions and the movement of free electrons.
When one end of a metallic rod is placed in a flame,the atoms at that end gain kinetic energy and vibrate more vigorously. These vibrations are passed on to neighboring atoms through intermolecular forces,effectively transferring energy along the rod.
In metals,the presence of a large number of free electrons significantly enhances this process. These electrons gain energy at the hot end and move rapidly toward the cooler end,colliding with atoms and transferring energy,which makes metals excellent thermal conductors.
Gases are generally poor thermal conductors,while liquids have thermal conductivities intermediate between solids and gases.

(N/A) double-walled flask or thermos bottle is an arrangement designed to minimize heat exchange between the surroundings and the system inside it. It consists of a cylindrical glass vessel with two walls.
$1$. $A$ silver coating is applied to both the inner and outer surfaces of these walls.
$2$. The inner silvered wall reflects thermal radiation back into the system,preventing heat loss.
$3$. The outer silvered wall reflects incoming thermal radiation from the surroundings,preventing heat gain.
$4$. The space between the two walls is evacuated (vacuum),which effectively eliminates heat transfer via conduction and convection.
$5$. The flask is sealed with an airtight insulating cork to prevent heat loss through evaporation or air circulation.
$6$. Consequently,the substance inside remains at its initial temperature (hot or cold) for an extended period.

(N/A) Heat transfer is the process of movement of thermal energy from a body at a higher temperature to a body at a lower temperature due to the temperature difference between them.
There are three modes of heat transfer:
$1$. Conduction: Heat transfer through a solid medium without the actual movement of particles.
$2$. Convection: Heat transfer through fluids (liquids and gases) involving the actual movement of particles.
$3$. Radiation: Heat transfer through electromagnetic waves.
Radiation is the mode of heat transfer that does not require any material medium for propagation.

(N/A) Heat transfer in solids occurs primarily through the process of $Conduction$.
In this process,heat energy is transferred from one part of a solid to another,or from one solid to another in physical contact,without the actual movement of the constituent particles (atoms or molecules) from their equilibrium positions.
Instead,the energy is transferred through molecular vibrations and the movement of free electrons (in metals),where higher-energy particles collide with their neighbors,transferring kinetic energy to them.

(N/A) Thermal steady state is a condition in heat transfer where the temperature at every point within a body remains constant over time.
In this state,although heat continues to flow through the material,the temperature at any given point does not change because the rate of heat entering the point is exactly equal to the rate of heat leaving it.
Mathematically,this implies that $\frac{dT}{dt} = 0$ for every point in the system,where $T$ is the temperature and $t$ is time.

(A) Snow consists of fine ice crystals with a large amount of air trapped between them. Since air is a very poor conductor of heat,the presence of these air pockets makes snow a much poorer conductor of heat compared to solid ice. Therefore,snow acts as an insulator.

(A) The temperature gradient is defined as the rate of change of temperature with respect to distance, given by the formula: $\frac{T_{H} - T_{C}}{L} = 80\,^{\circ}C/m$.
Here, $T_{H} = 30\,^{\circ}C$ (temperature of the hot end), $L = 0.5\,m$ (length of the rod), and $T_{C}$ is the temperature of the cold end.
Substituting the values into the equation:
$\frac{30 - T_{C}}{0.5} = 80$
$30 - T_{C} = 80 \times 0.5$
$30 - T_{C} = 40$
$T_{C} = 30 - 40$
$T_{C} = -10\,^{\circ}C$.
Therefore, the temperature of the cold end is $-10\,^{\circ}C$.

(N/A) The temperature gradient is defined as the rate of change of temperature with respect to distance in the direction of heat flow. Mathematically,it is expressed as $\frac{dT}{dx}$.
$1$. If the temperature difference is measured from the hot end towards the cold end (i.e.,in the direction of heat flow),the temperature decreases as distance increases. Thus,$\frac{dT}{dx}$ is negative.
$2$. If the temperature difference is measured from the cold end towards the hot end (i.e.,opposite to the direction of heat flow),the temperature increases as distance increases. Thus,$\frac{dT}{dx}$ is positive.

(C) The correct option is $C$.
The thermal resistance of a thin layer of paper is very low. When the cup is placed over a flame,the heat is transferred through the paper to the water inside.
Since water has a high specific heat capacity,it absorbs the heat effectively. As long as there is water in the cup,the temperature of the paper does not rise significantly above the temperature of the water (which is $100^{\circ} C$ at boiling point).
Because the temperature of the paper remains below its ignition (burning) temperature,the paper cup does not catch fire.

(D) The correct option is $D$.
When ice is crushed,the total surface area of the ice that comes into contact with the surrounding air or the contents of the cooler increases significantly.
According to the principles of heat transfer,the rate of heat exchange is directly proportional to the surface area in contact.
Therefore,increasing the surface area by crushing the ice accelerates the rate of heat absorption from the surroundings,thereby speeding up the cooling process.

(B) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K \Delta T}{L \rho x}$,where $K$ is thermal conductivity,$L$ is latent heat,$\rho$ is density,and $\Delta T$ is the temperature difference.
This implies $x \, dx = \frac{K \Delta T}{L \rho} dt$.
Integrating both sides,we get $t \propto x^2$.
Let $t_1$ be the time to form the first $1 \, cm$ $(x_1 = 1 \, cm)$ and $t_2$ be the time to form the next $1 \, cm$ (total thickness $x_2 = 2 \, cm$).
The time taken to reach thickness $x$ is $t = C x^2$.
For $x = 1 \, cm$,$t_1 = C(1)^2 = 24 \, h$.
For $x = 2 \, cm$,total time $t_{total} = C(2)^2 = 4C = 4 \times 24 = 96 \, h$.
The time taken to form the next $1 \, cm$ is $t_2 = t_{total} - t_1 = 96 - 24 = 72 \, h$.

(C) The phenomenon where a temperature difference is converted into electrical energy is known as the $Seebeck$ effect.
To maximize the efficiency of thermoelectric materials,we require a high $Seebeck$ coefficient,high electrical conductivity (to minimize Joule heating losses),and low thermal conductivity (to maintain the temperature gradient across the material).
Therefore,the material should have low thermal conductivity and high electrical conductivity.

(D) The rate of ice formation is given by the relation $t \propto (x_2^2 - x_1^2)$,where $x$ is the thickness of the ice layer.
For the first $3 \ mm$ $(x_1 = 3 \ mm)$,the time taken is $t_1 = 1 \ hour$.
For the next $6 \ mm$,the total thickness becomes $x_2 = 3 \ mm + 6 \ mm = 9 \ mm$.
The time taken for the total thickness of $9 \ mm$ is $T$.
The time taken for the next $6 \ mm$ is $t_2 = T - t_1$.
Using the proportionality: $\frac{t_1}{T} = \frac{x_1^2}{x_2^2} = \frac{3^2}{9^2} = \frac{9}{81} = \frac{1}{9}$.
So,$T = 9 \times t_1 = 9 \times 1 = 9 \ hours$.
The time taken for the next $6 \ mm$ is $t_2 = 9 - 1 = 8 \ hours$.

(B) The temperature gradient is defined as the rate of change of temperature with respect to distance,given by $T_g = \frac{T_h - T_c}{L}$,where $T_h$ is the temperature of the hotter end,$T_c$ is the temperature of the cooler end,and $L$ is the length of the rod.
Given:
Length $L = 75 \ cm = 0.75 \ m$
Temperature gradient $T_g = 40^{\circ} C/m$
Cooler end temperature $T_c = 10^{\circ} C$
Substituting the values into the formula:
$40 = \frac{T_h - 10}{0.75}$
$T_h - 10 = 40 \times 0.75$
$T_h - 10 = 30$
$T_h = 30 + 10 = 40^{\circ} C$
Therefore,the temperature of the hotter end is $40^{\circ} C$.

(A) At steady state,the temperature at different parts of the rod remains constant and is independent of time. The temperature gradient,which is the ratio of the temperature difference to the distance between two points,is constant throughout the rod.
Let $T_C$ be the temperature at a point $C$ located $9 \ cm$ from end $A$.
The temperature gradient is given by:
$\frac{T_A - T_C}{x_C - x_A} = \frac{T_A - T_B}{L}$
Substituting the given values:
$\frac{100 - T_C}{9} = \frac{100 - 0}{20}$
$\frac{100 - T_C}{9} = \frac{100}{20}$
$100 - T_C = 5 \times 9$
$100 - T_C = 45$
$T_C = 100 - 45 = 55^{\circ} C$
Therefore,the temperature at the point is $55^{\circ} C$.

(C) The space between the two walls of a thermos bottle is evacuated to prevent heat transfer due to conduction and convection.
However,heat transfer by radiation does not require a material medium and can occur through a vacuum.
Therefore,the vacuum between the walls does not inhibit heat transfer by radiation.
Thus,the Assertion is true,but the Reason is false.