Question 121:
moderateThe 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., $4.0 text{ J/s}$.
Question 122:
moderateA 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 $frac{Q}{t} = frac{K A Delta T}{x}$
$frac{m L}{t} = frac{K A (100 - 0)}{x}$
$K = frac{m L x}{t A Delta T} = frac{4.8 times 3.36 times 10^5 times 0.1}{3600 times 0.36 times 100} = 1.24 text{ J/m/s/}^circ text{C}$
Question 123:
moderateA 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 $Rightarrow pi r^2 l = pi (r/2)^2 l' Rightarrow l' = 4l$
$Q' = frac{K (pi (r/2)^2) Delta T}{4l} t = frac{1}{16} frac{K (pi r^2) Delta T}{l} t = frac{Q}{16}$
Question 124:
moderateThe 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 $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$
Question 125:
moderateConsider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities $K$ and $2K$, respectively. The equivalent thermal conductivity of the slab is:
(2003)
For compound slab in series, $K_{eq} = frac{l_1 + l_2}{frac{l_1}{K_1} + frac{l_2}{K_2}}$
$K_{eq} = frac{l + l}{frac{l}{K} + frac{l}{2K}} = frac{2l}{frac{3l}{2K}} = frac{4}{3} K$
Question 126:
moderateConsider two rods of same length and different specific heats ($S_1$, $S_2$), conductivities ($K_1$, $K_2$) and area of cross-sections ($A_1$, $A_2$) and both having temperature $T_1$ and $T_2$ at their ends. If rate of loss of heat due to conduction is equal, then
(2002)
Rate of heat loss $frac{dQ}{dt} = frac{K A (T_1 - T_2)}{l}$
Given $(frac{dQ}{dt})_1 = (frac{dQ}{dt})_2 Rightarrow frac{K_1 A_1 (T_1 - T_2)}{l} = frac{K_2 A_2 (T_1 - T_2)}{l}$
$Rightarrow K_1 A_1 = K_2 A_2$
Question 127:
moderateA cylindrical rod having temperature $T_1$ and $T_2$ at its ends. The rate of flow of heat $Q_1 text{ cal/sec}$. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_2$ will be:
(2001)
$Q_1 = frac{K A Delta T}{l} = frac{K (pi r^2) Delta T}{l}$
When dimensions are doubled, $r' = 2r$, $l' = 2l$
$Q_2 = frac{K (pi (2r)^2) Delta T}{2l} = frac{4 K (pi r^2) Delta T}{2l} = 2 Q_1$
Question 128:
moderateGravitational 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.
Question 129:
Heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio $1 : 2$ and the lengths in the ratio $2 : 1$. If the temperature difference between the ends is same, then ratio of the rate of flow of heat through them will be: (1995)
Rate of heat flow $H = \frac{KA\Delta T}{L} \propto \frac{r^2}{L}$. Ratio $H_1 / H_2 = (r_1/r_2)^2 \times (L_2/L_1) = (1/2)^2 \times (1/2) = 1/4 \times 1/2 = 1/8$.
Question 130:
moderateA 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 $t' = \frac{13}{5}t$.