Mercury thermometer can be used to measure temperature upto: (1992)
The boiling point of mercury is approximately $356.7^\circ\text{C}$. Therefore, a standard mercury thermometer can measure temperatures up to around $360^\circ\text{C}$.
A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers $140^\circ\text{F}$. What is the fall in temperature as registered by the centigrade thermometer? (1990)
Initial temperature of boiling water is $212^\circ\text{F}$. Fall in Fahrenheit $= 212^\circ\text{F} - 140^\circ\text{F} = 72^\circ\text{F}$. Using $\Delta C = \frac{5}{9} \Delta F$, fall in Celsius $= \frac{5}{9} \times 72 = 40^\circ\text{C}$.
The value of coefficient of volume expansion of glycerin is $5 \times 10^{-4}\text{ /K}$. The fractional change in the density of glycerin for a rise of $40^\circ\text{C}$ in its temperature, is: (2015 Re)
The fractional change in density is approximately given by $\frac{\Delta \rho}{\rho} = \gamma \Delta T$. Substituting the values: $\frac{\Delta \rho}{\rho} = 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$.
Which of the following rods, (given radius $r$ and length $l$) each made of the same material and whose ends are maintained at the same temperature will conduct most heat? (2005)
The rate of heat conduction is $H = \frac{KA\Delta T}{l} = \frac{K(\pi r^2)\Delta T}{l}$. Since material and $\Delta T$ are same, $H \propto \frac{r^2}{l}$. This ratio is maximum for $r = 2r_0$ and $l = l_0$ (ratio $\propto 4$).
The quantities of heat required to raise the temperature of two solid copper spheres of radii $r_1$ and $r_2$ ($r_1 = 1.5 r_2$) through $1\text{ K}$ are in the ratio: (2020)
Heat required $Q = mc\Delta T = (\frac{4}{3}\pi r^3 \rho)c\Delta T$. For the same material and $\Delta T$, $Q \propto r^3$. Ratio $= (\frac{r_1}{r_2})^3 = (1.5)^3 = (\frac{3}{2})^3 = \frac{27}{8}$.
A piece of ice falls from a height $h$ so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of $h$ is [Latent heat of ice is $3.4 \times 10^5\text{ J/kg}$ and $g = 10\text{ N/kg}$]: (2016 – I)
Energy absorbed by ice $= \frac{1}{4} mgh$. This energy melts the ice, so $\frac{1}{4} mgh = mL$. Substituting values: $h = \frac{4L}{g} = \frac{4 \times 3.4 \times 10^5}{10} = 13.6 \times 10^4\text{ m} = 136\text{ km}$.
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at $100^\circ\text{C}$, while the other one is at $0^\circ\text{C}$. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is:
(2016 – II)
By conservation of energy, $\int_{T_f}^{100} C(T)dT = \int_{0}^{T_f} C(T)dT$. Since $C(T)$ is greater at higher temperatures, the change in temperature for the hotter body will be less than that for the colder body. Thus, $T_f > 50^\circ\text{C}$.
Steam at $100^\circ\text{C}$ is passed into $20\text{ g}$ of water at $10^\circ\text{C}$. When water acquires a temperature of $80^\circ\text{C}$, the mass of water present will be: [Take specific heat of water $= 1\text{ cal /g }^\circ\text{C}$ and latent heat of steam $= 540\text{ cal g}^{-1}$]: (2014)
Heat gained by water $= 20 \times 1 \times (80 - 10) = 1400\text{ cal}$. Heat lost by $m$ grams of steam $= m \times 540 + m \times 1 \times (100 - 80) = 560m$. Equating them: $560m = 1400 \Rightarrow m = 2.5\text{ g}$. Total mass $= 20 + 2.5 = 22.5\text{ g}$.
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 $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$
Consider 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, $K_{eq} = frac{l_1 + l_2}{frac{l_1}{K_1} + frac{l_2}{K_2}}$
$K_{eq} = frac{l + l}{frac{l}{K} + frac{l}{2K}} = frac{2l}{frac{3l}{2K}} = frac{4}{3} K$