The unit of thermal conductivity is :
(2019)
Thermal conductivity $K = frac{Delta Q cdot x}{A cdot Delta T cdot t}$
Unit $= frac{text{J} cdot text{m}}{text{m}^2 cdot text{K} cdot text{s}} = frac{text{W}}{text{m K}} = text{W m}^{-1} text{K}^{-1}$
The two ends of a metal rod are maintained at temperatures $100^circ text{C}$ and $110^circ text{C}$. The rate of heat flow in the rod is found to be $4.0 text{ J/s}$. If the ends are maintained at temperatures $200^circ text{C}$ and $210^circ text{C}$, the rate of heat flow will be:
(2015)
Rate of heat flow $frac{dQ}{dt} = frac{K A Delta T}{L}$
Since $Delta T$ is same ($10^circ text{C}$) in both cases, the rate of heat flow will remain same i.e., $4.0 text{ J/s}$.
A slab of stone of area $0.36 text{ m}^2$ and thickness $0.1 text{ m}$ is exposed on the lower surface to steam at $100^circ text{C}$. A block of ice at $0^circ text{C}$ rests on the upper surface of the slab. In one hour $4.8 text{ kg}$ of ice is melted. The thermal conductivity of slab is: (Given latent heat of fusion of ice $= 3.36 times 10^5 text{ J/kg}$)
(2012 Mains)
Heat transferred $frac{Q}{t} = frac{K A Delta T}{x}$
$frac{m L}{t} = frac{K A (100 - 0)}{x}$
$K = frac{m L x}{t A Delta T} = frac{4.8 times 3.36 times 10^5 times 0.1}{3600 times 0.36 times 100} = 1.24 text{ J/m/s/}^circ text{C}$
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time $t$. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time $t$?
(2010 Pre)
$Q = frac{K A Delta T}{l} t = frac{K (pi r^2) Delta T}{l} t$
Volume is constant $Rightarrow pi r^2 l = pi (r/2)^2 l' Rightarrow l' = 4l$
$Q' = frac{K (pi (r/2)^2) Delta T}{4l} t = frac{1}{16} frac{K (pi r^2) Delta T}{l} t = frac{Q}{16}$
The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ ($T_1 > T_2$). The rate of heat transfer, $frac{dQ}{dt}$ through the rod in steady state is given by:
(2009)
Rate of heat transfer $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$
Gravitational force is required for: (2000)
Convection involves the macroscopic movement of fluid which relies on density differences. These differences in density lead to buoyant forces, which require gravity to operate.
A cup of coffee cools from $90^\circ\text{C}$ to $80^\circ\text{C}$ in $t$ minutes, when the room temperature is $20^\circ\text{C}$. The time taken by a similar cup of coffee to cool from $80^\circ\text{C}$ to $60^\circ\text{C}$ at a room temperature same at $20^\circ\text{C}$ is: (2021)
Using average form of Newton's law of cooling: $\frac{90-80}{t} = K(\frac{90+80}{2}-20) \Rightarrow \frac{10}{t} = K(65)$. For second case: $\frac{80-60}{t'} = K(\frac{80+60}{2}-20) \Rightarrow \frac{20}{t'} = K(50)$. Dividing the two equations yields $t' = \frac{13}{5}t$.
A body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next $10$ minutes will be: (2016 – II)
By Newton's law of cooling: $\frac{3T-2T}{10} = K(\frac{3T+2T}{2}-T) \Rightarrow \frac{T}{10} = K(1.5T)$. For next 10 mins: $\frac{2T-T'}{10} = K(\frac{2T+T'}{2}-T)$. Substituting $K = \frac{1}{15}$, we get $2T - T' = \frac{1}{15}(0.5T' + T) \times 10$. Solving gives $T' = \frac{3}{2}T$.
Certain quantity of water cools from $70^\circ\text{C}$ to $60^\circ\text{C}$ in the first $5$ minutes and to $54^\circ\text{C}$ in the next $5$ minutes. The temperature of the surroundings is: (2014)
Using Newton's law of cooling: $\frac{70-60}{5} = K(65-T_s) \Rightarrow 2 = K(65-T_s)$ and $\frac{60-54}{5} = K(57-T_s) \Rightarrow 1.2 = K(57-T_s)$. Dividing gives $\frac{2}{1.2} = \frac{65-T_s}{57-T_s} \Rightarrow 5(57-T_s) = 3(65-T_s)$. Solving for $T_s$, we get $T_s = 45^\circ\text{C}$.
A beaker full of hot water is kept in a room. If it cools from $80^\circ\text{C}$ to $75^\circ\text{C}$ in $t_1$ minutes, from $75^\circ\text{C}$ to $70^\circ\text{C}$ in $t_2$ minutes and from $70^\circ\text{C}$ to $65^\circ\text{C}$ in $t_3$ minutes, then: (1995)
According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and surroundings. As the water cools, the temperature difference decreases, slowing the rate of cooling. Thus, successive intervals take more time, giving $t_1 < t_2 < t_3$.