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 $= (\frac{r_1}{r_2})^3 = (1.5)^3 = (\frac{3}{2})^3 = \frac{27}{8}$.
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 $= 20 + 2.5 = 22.5\text{ g}$.
When $1 text{ kg}$ of ice at $0^circ text{C}$ melts to water at $0^circ text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 text{ Cal}/^circ text{C}$, is:
(2011 Pre)
$10 text{ gm}$ of ice cubes at $0^circ text{C}$ are released in a tumbler (water equivalent $55 text{ g}$) at $40^circ text{C}$. Assuming that negligible heat is taken from the surroundings the temperature of water in the tumbler becomes nearly ($L = 80 text{ cal/g}$):
(1988)