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 $= 20 + 2.5 = 22.5\text{ g}$.
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, $K_{eq} = frac{l_1 + l_2}{frac{l_1}{K_1} + frac{l_2}{K_2}}$
$K_{eq} = frac{l + l}{frac{l}{K} + frac{l}{2K}} = frac{2l}{frac{3l}{2K}} = frac{4}{3} K$
Consider 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 $(frac{dQ}{dt})_1 = (frac{dQ}{dt})_2 Rightarrow frac{K_1 A_1 (T_1 - T_2)}{l} = frac{K_2 A_2 (T_1 - T_2)}{l}$
$Rightarrow K_1 A_1 = K_2 A_2$
A 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, $r' = 2r$, $l' = 2l$
$Q_2 = frac{K (pi (2r)^2) Delta T}{2l} = frac{4 K (pi r^2) Delta T}{2l} = 2 Q_1$
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)
Entropy change $Delta S = frac{Delta Q}{T} = frac{m cdot L}{T} = frac{1000 cdot 80}{273} = 293 text{ cal/K}$
Thermal capacity of $40 text{ g}$ of aluminum ($s = 0.2 text{ cal/g K}$) is:
(1990)
Thermal capacity $= ms = 40 cdot 0.2 = 8 text{ cal/K} = 8 cdot 4.2 text{ J/K} = 33.6 text{ J/K}$
$10 text{ gm}$ 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)
Heat lost by tumbler = Heat gained by ice
$55 cdot (40 - T) = 10 cdot 80 + 10 cdot T$
$2200 - 55T = 800 + 10T Rightarrow 65T = 1400 Rightarrow T approx 21.5^circ text{C} approx 22^circ text{C}$