Heat Conduction in One Dimension
By In general, the temperature at a point (x) varies with time (t) i.e., it is a function $T(x,t)$. In the steady state, the temperature depends on x but not on time t i.e., $T(x,t)=T(x)$. In this state, the heat that reaches any cross-section is transmitted to the next without accumulation. The discussion in this article is limited conduction heat transfer in the steady state.
The rate of heat transfer through a rod of length x and cross-section area A whose two ends are maintained at a temperature T1 and T2 is given by \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{KA}{x}(T_1-T_2) \end{align} where K is the thermal conductivity of the material of the rod. The SI unit of thermal conductivity is W/(m-K). The dimensional formula of the thermal conductivity is MLT-3K-1. Note that heat is transferred from the end at higher temperature T1 to the end at the lower temperature T2.
The quantity dQ/dt is also called heat current or the rate of flow of heat. The quantity $(T_1-T_2)/x$ (and its differential form dT/dx) is called temperature gradient. The thermal resistance of a body is defined as the ratio of temperature difference to the heat current \begin{align} R=\frac{\Delta T}{\mathrm{d}Q/\mathrm{d}t}=\frac{x}{KA}. \end{align} In electrical circuits, equivalents of heat current, temperature difference and thermal resistance is electric current, potential difference and electrical resistance, respectively.
Consider two rods of thermal resistances $R_1$ and $R_2$. The effective thermal resistance of a system of two rods connected in series is given by \begin{align} R_s=R_1+R_2 \end{align} If these rods are of same physical dimensions and of thermal conductivities $K_1$ and $K_2$ then effective thermal conductivity of the system is \begin{align} K_s=\frac{K_1K_2}{K_1+K_2}. \end{align} The effective thermal resistance of a system of two rods connected in parallel is given by \begin{align} R_p=\frac{R_1R_2}{R_1+R_2}. \end{align} If rods is parrallel are of same physical dimensions but thermal conductivities $K_1$ and $K_2$ then effective thermal conductivity of the system is \begin{align} K_p=K_1+K_2. \end{align}
In general, a substance that is a good conductor of heat is also a good conductor of electricity. At a given temperature T, the ratio of thermal conductivity (K) to the electrical conductivity (σ) is constant i.e., K/σT= L (a constant). This is known as Wiedemann-Franz Law.
Solved Problems on Conduction Heat Transfer from IIT JEE
- The end Q and R of two thin wires, PQ and RS, are soldered (IIT JEE 2016)
- Two blocks connected in series and parallel ( IIT JEE 2013)
- Two rods connected in series and parallel (IIT JEE 2014)
- Two conducting cylinders of equal length but different radii connected in series (IIT JEE 2018)
- Temperature at the junction of the three identical rods (IIT JEE 2001)
- Three rods forms an isosceles triangle. Temperature at the vertex? (IIT JEE 1995)
- Effective thermal conductivity of two concentric cylinders (IIT JEE 1988)
- Evaporation of water and melting of ice kepts at two ends of a metal rod AB of length 10x (IIT JEE 2009)
- MCQ on a composite block is made of slabs A, B, C, D, and E (IIT JEE 2011)
- A point source of heat of power P kept at the centre of a spherical shell (IIT JEE 1991)
- A double pane window used for insulating a room thermally from outside (IIT JEE 1997)
- An electric heater is used in a room of total wall area 137 m2 ( IIT JEE 1986)
Questions on Conduction Heat Transfer
Question 1: For cooking the food, which of the following type of utensil is most suitable
Question 2: Two conducting walls of thickness d1 and d2, and thermal conductivity K1 and K2 are joined together. If temperatures on the outside surfaces are T1 and T2 then the temperature of common surface is
Question 3: Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at -5.0°C and the bottom of the pond at 4°C. If the total depth of ice + water is 1.4 m, then he thickness of ice layer is (Assume that the thermal conductivities of ice and water are 0.40 and 0.12 cal/(m-°C), respectively.)
