Newton's Law of Cooling Questions in English

(A) According to Newton's Law of Cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surroundings: $R \propto (T - T_s)$.
Given:
Temperature of body $A$ $(T_1)$ = $100^{\circ}C$
Temperature of body $B$ $(T_2)$ = $80^{\circ}C$
Surrounding temperature $(T_s)$ = $40^{\circ}C$
The rate of cooling for body $A$ is $R_1 = k(T_1 - T_s) = k(100 - 40) = 60k$.
The rate of cooling for body $B$ is $R_2 = k(T_2 - T_s) = k(80 - 40) = 40k$.
Therefore,the ratio $R_1 : R_2 = \frac{60k}{40k} = \frac{60}{40} = \frac{3}{2}$.

(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d\theta}{dt} = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.
Since the spheres are of the same material,the emissivity $e$,specific heat $s$,and density $\rho$ are the same.
Mass $m = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3$ and Surface Area $A = 4 \pi r^2$.
Substituting these into the formula: $\frac{d\theta}{dt} \propto \frac{A}{m} \propto \frac{r^2}{r^3} \propto \frac{1}{r}$.
Therefore,the ratio of the rates of cooling is $\frac{(d\theta/dt)_1}{(d\theta/dt)_2} = \frac{r_2}{r_1} = \frac{2r}{r} = 2:1$.

(D) According to Newton's Law of Cooling,the rate of heat loss is proportional to the temperature difference between the body and its surroundings: $\frac{dQ}{dt} = k(\theta - \theta_0)$.
Given: $\theta_0 = 20^{\circ}C$.
For the first case: $\theta_1 = 80^{\circ}C$ and $\frac{dQ_1}{dt} = 60 \, cal/sec$.
Substituting these values: $60 = k(80 - 20) = 60k$,which implies $k = 1 \, cal/(sec \cdot ^{\circ}C)$.
For the second case: $\theta_2 = 40^{\circ}C$.
Calculating the rate of heat loss: $\frac{dQ_2}{dt} = k(40 - 20) = 1 \times 20 = 20 \, cal/sec$.

(C) According to Newton's law of cooling,the rate of heat loss is given by $\frac{dQ}{dt} = mc \frac{d\theta}{dt}$.
Since the conditions are identical,the rate of heat loss $\frac{dQ}{dt}$ is the same for both liquids.
Therefore,$m_1 s_1 \frac{d\theta}{t_1} = m_2 s_2 \frac{d\theta}{t_2}$.
Since the volumes are equal $(V_1 = V_2 = V)$,we can write $m = \rho V$.
Substituting this,we get $\rho_1 V s_1 / t_1 = \rho_2 V s_2 / t_2$.
This simplifies to $\frac{\rho_1}{\rho_2} = \frac{s_2}{s_1} \times \frac{t_1}{t_2}$.
Given $s_1 : s_2 = 3 : 4$,so $s_2 / s_1 = 4/3$.
Given $t_1 = 324 \; sec$ and $t_2 = 810 \; sec$.
Ratio $\frac{\rho_1}{\rho_2} = \frac{4}{3} \times \frac{324}{810} = \frac{4}{3} \times \frac{2}{5} = \frac{8}{15}$.

(C) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the average temperature of the body and the surrounding temperature.
Mathematically,$\frac{d\theta}{dt} \propto \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right)$.
Since the average temperature of the water decreases as it cools,the rate of cooling decreases over time.
For the given intervals:
Average temperature $1$: $\frac{75+70}{2} = 72.5^{\circ}C$
Average temperature $2$: $\frac{70+65}{2} = 67.5^{\circ}C$
Average temperature $3$: $\frac{65+60}{2} = 62.5^{\circ}C$
Since the average temperature is highest for the first interval,the rate of cooling is fastest,meaning $T_1$ is the smallest time.
Therefore,$T_1 < T_2 < T_3$.

(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_s)$.
For the first interval: $\frac{90 - 60}{5} = K\left( \frac{90 + 60}{2} - 20 \right) \Rightarrow 6 = K(75 - 20) \Rightarrow 6 = 55K \Rightarrow K = \frac{6}{55}$.
For the second interval: $\frac{60 - 30}{t} = K\left( \frac{60 + 30}{2} - 20 \right) \Rightarrow \frac{30}{t} = \frac{6}{55}(45 - 20) \Rightarrow \frac{30}{t} = \frac{6}{55} \times 25$.
$\frac{30}{t} = \frac{6 \times 5}{11} \Rightarrow \frac{30}{t} = \frac{30}{11} \Rightarrow t = 11 \ min$.

(B) According to Stefan-Boltzmann Law,the rate of heat loss is given by $dQ/dt = \sigma A (T^4 - T_0^4)$.
For a small temperature difference,the rate of cooling $d\theta/dt$ is proportional to the rate of heat loss divided by the heat capacity $(ms)$:
$d\theta/dt = \frac{\sigma A (T^4 - T_0^4)}{ms}$.
Since $m = \rho V = \rho (\frac{4}{3}\pi r^3)$ and $A = 4\pi r^2$,we have:
$d\theta/dt \propto \frac{r^2}{\rho r^3 s} \propto \frac{1}{\rho r}$.
Given the ratios $r_1:r_2 = 1:2$ and $\rho_1:\rho_2 = 2:1$,the ratio of the rate of cooling is:
$\frac{(d\theta/dt)_1}{(d\theta/dt)_2} = \frac{\rho_2 r_2}{\rho_1 r_1} = \frac{1}{2} \times \frac{2}{1} = 1:1$.

(A) According to Newton's Law of Cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-dT/dt = k(T - T_S)$.
Here,$T$ is the average temperature of the body and $T_S$ is the temperature of the surroundings.
For a small temperature drop $\Delta T = 10 \, ^\circ\text{C}$,the time taken $t$ is given by $t \propto 1 / (T_{avg} - T_S)$.
$1$. For $100 \, ^\circ\text{C}$ to $90 \, ^\circ\text{C}$,$T_{avg} = 95 \, ^\circ\text{C}$,so $t_1 \propto 1 / (95 - T_S)$.
$2$. For $90 \, ^\circ\text{C}$ to $80 \, ^\circ\text{C}$,$T_{avg} = 85 \, ^\circ\text{C}$,so $t_2 \propto 1 / (85 - T_S)$.
$3$. For $80 \, ^\circ\text{C}$ to $70 \, ^\circ\text{C}$,$T_{avg} = 75 \, ^\circ\text{C}$,so $t_3 \propto 1 / (75 - T_S)$.
Since the average temperature $T_{avg}$ decreases in each step,the denominator $(T_{avg} - T_S)$ decreases,which means the time $t$ increases.
Therefore,$t_1 < t_2 < t_3$.

(C) According to Newton's law of cooling,$\frac{dT}{dt} = -K(T - T_{room})$.
For the first interval: $\frac{60 - 50}{10} = K\left( \frac{60 + 50}{2} - 25 \right) \implies 1 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30}$.
For the second interval,let the final temperature be $\theta$: $\frac{50 - \theta}{10} = K\left( \frac{50 + \theta}{2} - 25 \right)$.
Substituting $K = \frac{1}{30}$: $\frac{50 - \theta}{10} = \frac{1}{30} \left( \frac{50 + \theta - 50}{2} \right) = \frac{1}{30} \left( \frac{\theta}{2} \right) = \frac{\theta}{60}$.
$6(50 - \theta) = \theta \implies 300 - 6\theta = \theta \implies 7\theta = 300 \implies \theta = \frac{300}{7} \approx 42.85^{\circ}C$.

(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{\theta_1 - \theta_2}{t} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right)$,where $\theta_0$ is the room temperature.
In the first case:
$\frac{80 - 70}{30} = K \left( \frac{80 + 70}{2} - \theta_0 \right) \Rightarrow \frac{10}{30} = K(75 - \theta_0) \Rightarrow \frac{1}{3} = K(75 - \theta_0)$ --- $(1)$
In the second case:
$\frac{60 - 50}{70} = K \left( \frac{60 + 50}{2} - \theta_0 \right) \Rightarrow \frac{10}{70} = K(55 - \theta_0) \Rightarrow \frac{1}{7} = K(55 - \theta_0)$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{1/3}{1/7} = \frac{K(75 - \theta_0)}{K(55 - \theta_0)} \Rightarrow \frac{7}{3} = \frac{75 - \theta_0}{55 - \theta_0}$
Cross-multiplying:
$7(55 - \theta_0) = 3(75 - \theta_0)$
$385 - 7\theta_0 = 225 - 3\theta_0$
$385 - 225 = 7\theta_0 - 3\theta_0$
$160 = 4\theta_0$
$\theta_0 = 40^{\circ} C$.

(B) According to Newton's Law of Cooling: $\frac{dT}{dt} = -K(T_{avg} - T_s)$.
For the first case: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}{2} - 30 \right) \Rightarrow \frac{0.1}{5} = K(49.95 - 30) \Rightarrow 0.02 = K(19.95) \dots (i)$.
For the second case: $\frac{40.0 - 39.9}{t} = K \left( \frac{40.0 + 39.9}{2} - 30 \right) \Rightarrow \frac{0.1}{t} = K(39.95 - 30) \Rightarrow \frac{0.1}{t} = K(9.95) \dots (ii)$.
Dividing $(i)$ by $(ii)$: $\frac{0.02}{0.1/t} = \frac{19.95}{9.95} \Rightarrow 0.2t = 2.005 \Rightarrow t \approx 10 \, s$.

(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the surrounding temperature.
For the first interval: $\frac{75 - 65}{2} = k \left( \frac{75 + 65}{2} - 30 \right) \implies 5 = k(70 - 30) \implies 5 = 40k \implies k = \frac{5}{40} = \frac{1}{8} \, \text{min}^{-1}$.
For the second interval: $\frac{55 - 45}{t} = k \left( \frac{55 + 45}{2} - 30 \right) \implies \frac{10}{t} = k(50 - 30) \implies \frac{10}{t} = 20k$.
Substituting $k = \frac{1}{8}$ into the second equation: $\frac{10}{t} = 20 \times \frac{1}{8} \implies \frac{10}{t} = 2.5 \implies t = \frac{10}{2.5} = 4 \, \text{min}$.

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the average temperature of the liquid and the surrounding temperature $T_0$.
For the first case:
$\frac{95 - 90}{30} = K \left( \frac{95 + 90}{2} - T_0 \right)$
$\frac{5}{30} = K (92.5 - T_0) \implies \frac{1}{6} = K (92.5 - T_0) \quad ---(1)$
For the second case:
$\frac{55 - 50}{70} = K \left( \frac{55 + 50}{2} - T_0 \right)$
$\frac{5}{70} = K (52.5 - T_0) \implies \frac{1}{14} = K (52.5 - T_0) \quad ---(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{1/6}{1/14} = \frac{92.5 - T_0}{52.5 - T_0}$
$\frac{14}{6} = \frac{92.5 - T_0}{52.5 - T_0} \implies \frac{7}{3} = \frac{92.5 - T_0}{52.5 - T_0}$
$7(52.5 - T_0) = 3(92.5 - T_0)$
$367.5 - 7T_0 = 277.5 - 3T_0$
$367.5 - 277.5 = 4T_0$
$90 = 4T_0$
$T_0 = 22.5^{\circ}C$

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = \frac{e \sigma A}{ms} (\theta - \theta_0)$.
Since $m = \rho V$,we have $\frac{d\theta}{dt} \propto \frac{A}{V}$.
For a cube of side $a$,$A = 6a^2$ and $V = a^3$,so $\frac{A}{V} = \frac{6}{a}$.
For a cube of side $2a$,$A' = 6(2a)^2 = 24a^2$ and $V' = (2a)^3 = 8a^3$,so $\frac{A'}{V'} = \frac{24}{8a} = \frac{3}{a}$.
Comparing the rates: $\frac{(d\theta/dt)'}{(d\theta/dt)} = \frac{A'/V'}{A/V} = \frac{3/a}{6/a} = \frac{1}{2}$.
Since the temperature range $(\theta_1 - \theta_2)$ is the same,the rate of cooling is inversely proportional to the time taken: $\frac{1/t'}{1/t} = \frac{1}{2}$.
Therefore,$t' = 2t$.

(C) According to Newton's Law of Cooling,the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings: $\frac{dQ}{dt} \propto (T - T_s)$.
In the first interval,the temperature drops from $100^{\circ}C$ to $80^{\circ}C$. The average temperature difference between the water and the surroundings is higher during this interval.
In the second interval,the temperature drops from $80^{\circ}C$ to $60^{\circ}C$. The average temperature difference between the water and the surroundings is lower during this interval.
Since the rate of cooling is higher when the temperature difference is larger,the time taken to cool by the same amount $(20^{\circ}C)$ will be less for the first interval.
Therefore,$T_1 < T_2$.

(B) According to Newton's Law of Cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundings: $-\frac{dT}{dt} = k(T - T_s)$.
As the temperature of the water decreases,the temperature difference $(T - T_s)$ between the water and the surroundings decreases.
Since the rate of cooling $(-\frac{dT}{dt})$ is proportional to this difference,the rate of cooling decreases over time.
Therefore,to lose the same amount of heat (a $5^{\circ}C$ drop),more time is required as the temperature approaches the ambient temperature.
Thus,$T_1 < T_2 < T_3$.

(A) According to Newton's law of cooling,the rate of heat loss $\frac{dQ}{dt}$ depends on the surface area and temperature difference. Since both spheres have the same radius and material,their surface areas are equal,so $\frac{dQ}{dt}$ is the same for both.
We know that $\frac{dQ}{dt} = mc \frac{dT}{dt}$,where $m$ is the mass,$c$ is the specific heat capacity,and $\frac{dT}{dt}$ is the rate of cooling.
Therefore,$\frac{dT}{dt} = \frac{1}{mc} \cdot \frac{dQ}{dt}$.
Since the material is the same,$c$ is constant. Because the hollow sphere has a smaller mass $(m)$ than the solid sphere,the rate of cooling $\frac{dT}{dt}$ will be higher for the hollow sphere.

(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the average temperature of the body and the surrounding temperature: $\frac{d\theta}{dt} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right)$.
For the first interval ($80^{\circ}C$ to $60^{\circ}C$ in $60 \, sec$):
$\frac{80 - 60}{60} = k \left( \frac{80 + 60}{2} - 30 \right) \implies \frac{20}{60} = k(70 - 30) \implies \frac{1}{3} = k(40) \implies k = \frac{1}{120} \, sec^{-1}$.
For the second interval ($60^{\circ}C$ to $50^{\circ}C$ in $t \, sec$):
$\frac{60 - 50}{t} = k \left( \frac{60 + 50}{2} - 30 \right) \implies \frac{10}{t} = k(55 - 30) \implies \frac{10}{t} = k(25)$.
Substituting $k = \frac{1}{120}$:
$\frac{10}{t} = \frac{1}{120} \times 25 \implies \frac{10}{t} = \frac{25}{120} \implies \frac{10}{t} = \frac{5}{24}$.
$5t = 240 \implies t = 48 \, sec$.
Rounding to the nearest option,the answer is $50 \, sec$.

(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d\theta}{dt} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right)$,where $\theta_s$ is the surrounding temperature.
For the first interval ($60^oC$ to $50^oC$ in $10 \, min$):
$\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - \theta_s \right)$
$1 = K(55 - \theta_s) \quad \dots(i)$
For the second interval ($50^oC$ to $42^oC$ in $10 \, min$):
$\frac{50 - 42}{10} = K \left( \frac{50 + 42}{2} - \theta_s \right)$
$0.8 = K(46 - \theta_s) \quad \dots(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1}{0.8} = \frac{55 - \theta_s}{46 - \theta_s}$
$1.25 = \frac{55 - \theta_s}{46 - \theta_s}$
$1.25(46 - \theta_s) = 55 - \theta_s$
$57.5 - 1.25\theta_s = 55 - \theta_s$
$2.5 = 0.25\theta_s$
$\theta_s = 10^oC$.

(A) Let $T_s$ be the temperature of the surroundings.
According to $Newton's$ law of cooling,the rate of cooling is given by:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
For the first $5$ minutes:
$T_1 = 70^\circ C, T_2 = 60^\circ C, t = 5$ minutes.
$\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_s \right)$
$2 = K(65 - T_s)$ --- $(i)$
For the next $5$ minutes:
$T_1 = 60^\circ C, T_2 = 54^\circ C, t = 5$ minutes.
$\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_s \right)$
$1.2 = K(57 - T_s)$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2}{1.2} = \frac{65 - T_s}{57 - T_s}$
$\frac{5}{3} = \frac{65 - T_s}{57 - T_s}$
$5(57 - T_s) = 3(65 - T_s)$
$285 - 5T_s = 195 - 3T_s$
$2T_s = 90$
$T_s = 45^\circ C$

(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and the surroundings:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
For the first interval of $10$ minutes:
$T_1 = 3T, T_2 = 2T, T_s = T, t = 10$
$\frac{3T - 2T}{10} = K \left( \frac{3T + 2T}{2} - T \right)$
$\frac{T}{10} = K \left( 2.5T - T \right) = K(1.5T) \implies K = \frac{1}{15} \dots (i)$
For the next interval of $10$ minutes:
Let the final temperature be $T'$.
$T_1 = 2T, T_2 = T', T_s = T, t = 10$
$\frac{2T - T'}{10} = K \left( \frac{2T + T'}{2} - T \right)$
$\frac{2T - T'}{10} = K \left( \frac{T'}{2} \right) \dots (ii)$
Substituting $K = \frac{1}{15}$ into equation $(ii)$:
$\frac{2T - T'}{10} = \frac{1}{15} \left( \frac{T'}{2} \right)$
$\frac{2T - T'}{10} = \frac{T'}{30}$
$3(2T - T') = T'$
$6T - 3T' = T'$
$6T = 4T' \implies T' = \frac{6}{4}T = \frac{3}{2}T$

(A) According to Newton's Law of Cooling,the rate of heat loss is proportional to the temperature difference between the object and its surroundings,i.e.,$\frac{dQ}{dt} \propto (T - T_s)$.
In case of friend $A$,the cold milk is added immediately. This lowers the initial temperature of the tea mixture. Since the temperature difference between the mixture and the surroundings is smaller,the rate of heat loss is lower.
In case of friend $B$,the tea remains at a higher temperature for a longer duration before the cold milk is added. Because the temperature difference between the hot tea and the surroundings is larger,the rate of heat loss is higher.
Therefore,the tea in the cup of friend $A$ will remain hotter when the friend arrives.

(C) The rate of cooling $R_C$ is given by the formula: $R_C = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.
Since $m = \rho V$,where $\rho$ is density and $V$ is volume,we have $R_C = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{V \rho c}$.
For a sphere,$A = 4 \pi r^2$ and $V = \frac{4}{3} \pi r^3$,so $\frac{A}{V} = \frac{3}{r}$.
Thus,$R_C \propto \frac{1}{r} \propto \frac{1}{\text{Diameter}}$.
Given that the diameter of $A$ is half that of $B$ $(D_A = \frac{1}{2} D_B)$,the rate of cooling of $A$ will be twice that of $B$ $(R_{C,A} = 2 R_{C,B})$.

(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d\theta}{dt} = K(\theta_{avg} - \theta_s)$,where $\theta_{avg}$ is the average temperature of the body and $\theta_s$ is the surrounding temperature.
For the first case: $\frac{62 - 61}{T} = K \left( \frac{62 + 61}{2} - 30 \right) \implies \frac{1}{T} = K(61.5 - 30) = K(31.5)$ ...$(i)$
For the second case,cooling from $46^{\circ}C$ to $45^{\circ}C$ in time $T'$: $\frac{46 - 45}{T'} = K \left( \frac{46 + 45}{2} - 30 \right) \implies \frac{1}{T'} = K(45.5 - 30) = K(15.5)$ ...$(ii)$
Comparing $(i)$ and $(ii)$,since $31.5 > 15.5$,the rate of cooling is faster in the first case. Therefore,the time taken $T'$ for the second case must be greater than $T$.

(C) The rate of cooling is given by $\frac{dT}{dt} = \frac{kA(T - T_0)}{mc}$,where $m$ is the mass and $c$ is the specific heat capacity.
Since the mass $m$ and the surface area $A$ are the same for both,the rate of cooling is inversely proportional to the specific heat capacity $c$,i.e.,$\text{Rate of cooling} \propto \frac{1}{c}$.
Given that $c_{\text{oil}} < c_{\text{water}}$,it follows that $(\text{Rate of cooling})_{\text{oil}} > (\text{Rate of cooling})_{\text{water}}$.
This means that the oil cools down faster than the water.
Therefore,at any given time $t > 0$,the temperature of the oil will be lower than the temperature of the water.
Looking at the graph,curve $B$ shows a faster drop in temperature compared to curve $A$.
Thus,curve $B$ represents the cooling curve of oil,and curve $A$ represents the cooling curve of water.

(D) According to Newton's law of cooling, the rate of heat loss is $\frac{dQ}{dt} = k(T - T_{surr})$.
When the temperature is constant at $50^{\circ}C$, the heat supplied equals the heat lost: $10 = k(50 - 20) = 30k$. Thus, $k = \frac{1}{3} \ W/^{\circ}C$.
When the heater is switched off, the system cools. The average temperature during the cooling interval is $T_{avg} = \frac{35.1 + 34.9}{2} = 35^{\circ}C$.
The rate of heat loss at this average temperature is $\frac{dQ}{dt} = k(T_{avg} - T_{surr}) = \frac{1}{3}(35 - 20) = \frac{15}{3} = 5 \ J/s$.
In $1 \ minute$ $(60 \ seconds)$, the total heat lost is $Q = 5 \ J/s \times 60 \ s = 300 \ J$.
The change in temperature is $\Delta T = 35.1^{\circ}C - 34.9^{\circ}C = 0.2^{\circ}C$.
The heat capacity $C$ is given by $C = \frac{Q}{\Delta T} = \frac{300 \ J}{0.2^{\circ}C} = 1500 \ J/^{\circ}C$.

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $\frac{dT}{dt} = k(T - T_s)$,where $T_s$ is the surrounding temperature.
$CASE-1$: The body cools from $60^{\circ}C$ to $50^{\circ}C$ in $4 \text{ min}$. The average temperature is $\frac{60+50}{2} = 55^{\circ}C$.
$\frac{60-50}{4} = k(55 - T_s) \Rightarrow 2.5 = k(55 - T_s)$ $(1)$
$CASE-2$: The body cools from $40^{\circ}C$ to $30^{\circ}C$ in $8 \text{ min}$. The average temperature is $\frac{40+30}{2} = 35^{\circ}C$.
$\frac{40-30}{8} = k(35 - T_s) \Rightarrow 1.25 = k(35 - T_s)$ $(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{2.5}{1.25} = \frac{55 - T_s}{35 - T_s}$
$2 = \frac{55 - T_s}{35 - T_s}$
$70 - 2T_s = 55 - T_s$
$T_s = 15^{\circ}C$
Therefore,the temperature of the surrounding is $15^{\circ}C$.

(B) According to Newton's law of cooling,the rate of change of temperature is proportional to the temperature difference between the body and its surroundings:
$\frac{d\theta}{dt} = -k(\theta - \theta_0)$
Rearranging the terms for integration:
$\frac{d\theta}{\theta - \theta_0} = -k dt$
Integrating both sides:
$\int \frac{d\theta}{\theta - \theta_0} = \int -k dt$
$\ln(\theta - \theta_0) = -kt + C$
This equation is of the form $y = mx + c$,where $y = \ln(\theta - \theta_0)$,$x = t$,$m = -k$ (a negative slope),and $c$ is the intercept. This represents a straight line with a negative slope. Therefore,the correct graph is a straight line sloping downwards.

(C) According to Newton's law of cooling,the rate of loss of heat is proportional to the difference in temperature between the body and its surroundings,i.e.,$\frac{dT}{dt} = -k(T - \theta_0)$.
This differential equation leads to an exponential decay solution of the form $T(t) = \theta_0 + (\theta - \theta_0)e^{-kt}$.
As time $t$ increases,the temperature $T$ decreases exponentially and asymptotically approaches the ambient temperature $\theta_0$.
Graph $C$ correctly represents this exponential decay curve where the temperature starts at $\theta$ and approaches $\theta_0$ as $t \to \infty$.

(C) According to Newton's law of cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings:
$\frac{\theta_{1}-\theta_{2}}{t} = K \left( \frac{\theta_{1}+\theta_{2}}{2} - \theta_{s} \right)$
$Case-I$: Cooling from $62^{\circ}C$ to $50^{\circ}C$ in $10$ minutes with surrounding temperature $\theta_{s} = 26^{\circ}C$.
$\frac{62-50}{10} = K \left( \frac{62+50}{2} - 26 \right)$
$\frac{12}{10} = K (56 - 26)$
$1.2 = K \times 30$
$K = \frac{1.2}{30} = 0.04$
$Case-II$: Cooling from $50^{\circ}C$ to $\theta_{2}$ in the next $10$ minutes.
$\frac{50-\theta_{2}}{10} = K \left( \frac{50+\theta_{2}}{2} - 26 \right)$
$\frac{50-\theta_{2}}{10} = 0.04 \left( \frac{50+\theta_{2}-52}{2} \right)$
$50 - \theta_{2} = 0.4 \left( \frac{\theta_{2}-2}{2} \right)$
$50 - \theta_{2} = 0.2 (\theta_{2} - 2)$
$50 - \theta_{2} = 0.2\theta_{2} - 0.4$
$1.2\theta_{2} = 50.4$
$\theta_{2} = \frac{50.4}{1.2} = 42^{\circ}C$.

(B) According to Newton's law of cooling,the rate of heat loss is proportional to the temperature difference between the body and its surroundings: $P = K(T - T_s)$,where $K$ is a constant,$T$ is the temperature of the body,and $T_s$ is the temperature of the surroundings.
In the first case,the body is at $T_1 = 50^{\circ} C$ and the surroundings are at $T_s = 20^{\circ} C$. The heater provides $P_1 = 100 \ W$ to maintain the steady state.
$100 = K(50 - 20)$
$100 = K \times 30$
$K = \frac{100}{30} = \frac{10}{3} \ W/^{\circ} C$
In the second case,the body is at $T_2 = 35^{\circ} C$ and the surroundings are at $T_s = 20^{\circ} C$. Let the required power be $P_2$.
$P_2 = K(35 - 20)$
$P_2 = K \times 15$
Substituting the value of $K$:
$P_2 = \left(\frac{100}{30}\right) \times 15$
$P_2 = \frac{100}{2} = 50 \ W$.

(D) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the average temperature of the liquid and the surrounding temperature:
$\frac{\theta_1 - \theta_2}{t} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right)$
For the first interval ($50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes):
$\frac{50 - 45}{5} = K \left( \frac{50 + 45}{2} - \theta_0 \right) \implies 1 = K (47.5 - \theta_0) \quad ...(i)$
For the second interval ($45^{\circ}C$ to $41.5^{\circ}C$ in $5$ minutes):
$\frac{45 - 41.5}{5} = K \left( \frac{45 + 41.5}{2} - \theta_0 \right) \implies 0.7 = K (43.25 - \theta_0) \quad ...(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1}{0.7} = \frac{47.5 - \theta_0}{43.25 - \theta_0}$
$43.25 - \theta_0 = 0.7(47.5 - \theta_0)$
$43.25 - \theta_0 = 33.25 - 0.7\theta_0$
$0.3\theta_0 = 10$
$\theta_0 = \frac{10}{0.3} = 33.3^{\circ}C$

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $(T - T_s) = \Delta T$ is the temperature difference.
Given the initial rate of cooling is $R = k \Delta T_0$,we can find the rate constant $k = \frac{R}{\Delta T_0}$.
The temperature difference at any time $t$ is given by $\Delta T(t) = \Delta T_0 e^{-kt}$.
We want to find the time $t$ when $\Delta T(t) = \frac{\Delta T_0}{2}$.
Substituting this into the equation: $\frac{\Delta T_0}{2} = \Delta T_0 e^{-kt}$.
This simplifies to $\frac{1}{2} = e^{-kt}$,or $2 = e^{kt}$.
Taking the natural logarithm on both sides: $\ln(2) = kt$.
Therefore,$t = \frac{\ln(2)}{k}$.
Substituting $k = \frac{R}{\Delta T_0}$,we get $t = \frac{\ln(2) \Delta T_0}{R}$.

(D) According to Newton's Law of Cooling,the rate of heat loss is given by $\frac{dQ}{dt} = \sigma A (T - T_0)$,where $\sigma$ is a constant related to the surface nature,$A$ is the surface area,and $(T - T_0)$ is the temperature difference.
Since $dQ = mc \cdot dT$,we have $mc \frac{dT}{dt} = -\sigma A (T - T_0)$.
The rate of cooling is $\frac{-dT}{dt} = \frac{\sigma A}{mc} (T - T_0)$.
Substituting the values for a sphere: $A = 4 \pi r^2$ and $m = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3$.
Thus,$\frac{-dT}{dt} = \frac{\sigma (4 \pi r^2)}{\rho (\frac{4}{3} \pi r^3) c} (T - T_0)$.
Simplifying this,we get $\frac{-dT}{dt} \propto \frac{1}{\rho r c}$.

(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.
Since the rate of cooling $\frac{dT}{dt} = \frac{dQ/dt}{mc} = \frac{e \sigma A (T^4 - T_0^4)}{\rho V c}$,the rate of cooling is directly proportional to the surface area $A$ for a fixed volume $V$.
Among a sphere,a cube,and a disc of the same volume,the sphere has the minimum surface area.
Therefore,the sphere will have the lowest rate of cooling.

(C) According to Stefan's Law,the rate of cooling (rate of fall of temperature) is given by $\frac{d\theta}{dt} = \frac{\sigma A (T^4 - T_0^4)}{MS}$.
Since $M = V \rho = \frac{4}{3} \pi R^3 \rho$ and $A = 4 \pi R^2$,we have $\frac{d\theta}{dt} \propto \frac{A}{MS} = \frac{4 \pi R^2}{(\frac{4}{3} \pi R^3 \rho) S} = \frac{3}{R \rho S}$.
Therefore,the ratio of the rates of cooling is $\frac{(d\theta/dt)_1}{(d\theta/dt)_2} = \frac{R_2 \rho_2 S_2}{R_1 \rho_1 S_1}$.

(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = K(\theta_{avg} - \theta_0)$.
For the first interval: $\frac{80 - 60}{1} = K \left( \frac{80 + 60}{2} - 30 \right)$.
$20 = K(70 - 30) = 40K \implies K = \frac{20}{40} = 0.5 \text{ min}^{-1}$.
For the second interval: $\frac{60 - 50}{t} = K \left( \frac{60 + 50}{2} - 30 \right)$.
$\frac{10}{t} = 0.5 (55 - 30) = 0.5 \times 25 = 12.5$.
$t = \frac{10}{12.5} = 0.8 \text{ minutes}$.
Since $1 \text{ minute} = 60 \text{ seconds}$,$t = 0.8 \times 60 = 48 \text{ seconds}$.

(A) The rate of cooling $R = \frac{dT}{dt} = \frac{e \sigma A (T^4 - T_0^4)}{ms}$.
Since $m = \rho V$,we have $R = \frac{e \sigma A (T^4 - T_0^4)}{\rho V s}$.
For the same material,$e, \sigma, \rho, s$ are constant. Given the same surface area $A$ and temperature $T$,$R \propto \frac{1}{V}$.
For a sphere of radius $r$,$A = 4\pi r^2 \Rightarrow r = \sqrt{\frac{A}{4\pi}}$. Volume $V_s = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{A}{4\pi})^{3/2} = \frac{A^{3/2}}{6\sqrt{\pi}}$.
For a cube of side $a$,$A = 6a^2 \Rightarrow a = \sqrt{\frac{A}{6}}$. Volume $V_c = a^3 = (\frac{A}{6})^{3/2} = \frac{A^{3/2}}{6\sqrt{6}}$.
Ratio $\frac{R_s}{R_c} = \frac{V_c}{V_s} = \frac{A^{3/2} / 6\sqrt{6}}{A^{3/2} / 6\sqrt{\pi}} = \frac{\sqrt{\pi}}{\sqrt{6}} = \sqrt{\frac{\pi}{6}}$.
Thus,the ratio is $\sqrt{\frac{\pi}{6}} : 1$.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T$ is the temperature of the body,$T_s$ is the surrounding temperature,and $k$ is a constant.
For the first interval: $\frac{60 - 40}{10} = k \left( \frac{60 + 40}{2} - 20 \right)$.
$2 = k(50 - 20) = 30k$,so $k = \frac{2}{30} = \frac{1}{15}$.
For the next interval of $10$ minutes,let the final temperature be $T_f$.
$\frac{40 - T_f}{10} = k \left( \frac{40 + T_f}{2} - 20 \right)$.
Substituting $k = \frac{1}{15}$: $\frac{40 - T_f}{10} = \frac{1}{15} \left( \frac{40 + T_f - 40}{2} \right) = \frac{1}{15} \left( \frac{T_f}{2} \right) = \frac{T_f}{30}$.
Multiplying by $30$: $3(40 - T_f) = T_f$.
$120 - 3T_f = T_f \implies 4T_f = 120 \implies T_f = 30\,^{\circ}\text{C}$.

(B) The rate of cooling of a body is given by the formula:
$R = \frac{d\theta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$
Since the material is the same,the emissivity $\epsilon$,specific heat $c$,and density $\rho$ are constant.
$R \propto \frac{A}{m} = \frac{A}{\rho V} \propto \frac{A}{V}$
Given that the surface areas $A$ are equal,the rate of cooling $R$ is inversely proportional to the volume $V$ $(R \propto \frac{1}{V})$.
For a given surface area,the sphere has the maximum volume among all geometric shapes.
Therefore,$V_{\text{sphere}} > V_{\text{cube}}$.
Since $R \propto \frac{1}{V}$,it follows that $R_{\text{cube}} > R_{\text{sphere}}$.
Thus,the cube cools down faster than the sphere.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = \frac{e \sigma A_{surf}}{ms} (\theta^4 - \theta_0^4)$.
For small temperature differences,this simplifies to $\frac{d\theta}{dt} \propto \frac{1}{ms}(\theta - \theta_0)$.
Since the discs have equal mass $m$,equal surface area $A_{surf}$,and are under identical conditions,the rate of cooling $\frac{d\theta}{dt}$ is inversely proportional to the specific heat $s$ for a given temperature difference $(\theta - \theta_0)$.
From the graph,the slope of the line represents $\frac{d\theta/dt}{\theta - \theta_0} \propto \frac{1}{s}$.
Since the slope of line $A$ is greater than the slope of line $B$,the specific heat of $A$ must be less than the specific heat of $B$.

(B) According to Newton's law of cooling,the rate of change of temperature is proportional to the temperature difference between the body and its surroundings:
$\frac{-d\theta}{dt} = k(\theta - \theta_0)$
Rearranging the terms,we get:
$\frac{d\theta}{\theta - \theta_0} = -k dt$
Integrating both sides:
$\int \frac{d\theta}{\theta - \theta_0} = \int -k dt$
$\log_e(\theta - \theta_0) = -kt + C$
This equation is of the form $y = mx + c$,where $y = \log_e(\theta - \theta_0)$,$x = t$,$m = -k$ (a negative slope),and $c$ is the intercept.
Since the slope is negative,the graph is a straight line sloping downwards. This corresponds to Graph $B$.

(C) The rate of cooling $R_F = \frac{d\theta}{dt}$ is given by Newton's Law of Cooling as $R_F \propto \frac{1}{ms}$,where $m$ is the mass and $s$ is the specific heat capacity.
Since the spheres have the same radius,their volumes $V$ are equal. Mass $m = \rho V$,where $\rho$ is the density.
Thus,$R_F \propto \frac{1}{(\rho V)s} \propto \frac{1}{\rho s}$.
We need to calculate the product $\rho s$ for each sphere:
For $A$: $\rho s = 6 \times 2 = 12$
For $B$: $\rho s = 3 \times 5 = 15$
For $C$: $\rho s = 4 \times 4 = 16$
For $D$: $\rho s = 5 \times 6 = 30$
The rate of cooling is slowest when the product $\rho s$ is maximum.
Comparing the values $12, 15, 16, 30$,the maximum value is $30$,which corresponds to sphere $D$.

(D) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s$ is the surrounding temperature.
For the first interval: $\frac{80 - 60}{1} = K \left( \frac{80 + 60}{2} - 30 \right) \implies 20 = K(70 - 30) \implies 20 = 40K \implies K = 0.5\,min^{-1}$.
For the second interval: $\frac{60 - 50}{t} = K \left( \frac{60 + 50}{2} - 30 \right) \implies \frac{10}{t} = 0.5(55 - 30) \implies \frac{10}{t} = 0.5 \times 25 \implies \frac{10}{t} = 12.5$.
Solving for $t$: $t = \frac{10}{12.5} = 0.8\,min$.
Converting to seconds: $t = 0.8 \times 60 = 48\,sec$.

(B) According to Newton's law of cooling,the rate of loss of heat $(dQ/dt)$ of a body is directly proportional to the difference in temperature between the body and its surroundings,provided the temperature difference is small.
Mathematically,$-dT/dt = k(T - T_0)$,where $T$ is the temperature of the body,$T_0$ is the temperature of the surroundings,and $k$ is a constant.
Integrating this differential equation gives $T(t) = T_0 + (T_{initial} - T_0)e^{-kt}$.
This equation represents an exponential decay curve where the temperature $T$ asymptotically approaches the ambient temperature $T_0$ as time $t$ increases.
Therefore,the curve in option $B$ correctly represents this cooling process.

(D) According to Newton's law of cooling: $\frac{\theta_{1}-\theta_{2}}{t} = K \left( \frac{\theta_{1}+\theta_{2}}{2} - \theta_{0} \right)$.
For the first case: $\theta_{1} = 80\,^{\circ}C$,$\theta_{2} = 70\,^{\circ}C$,$t = 2\, \text{min} = 120\, \text{s}$,and $\theta_{0} = 30\,^{\circ}C$.
$\frac{80-70}{120} = K \left( \frac{80+70}{2} - 30 \right) \Rightarrow \frac{10}{120} = K(75 - 30) \Rightarrow \frac{1}{12} = 45K \Rightarrow K = \frac{1}{540}$.
For the second case: $\theta_{1} = 60\,^{\circ}C$,$\theta_{2} = 50\,^{\circ}C$,and $\theta_{0} = 30\,^{\circ}C$.
$\frac{60-50}{t} = K \left( \frac{60+50}{2} - 30 \right) \Rightarrow \frac{10}{t} = \frac{1}{540} (55 - 30) \Rightarrow \frac{10}{t} = \frac{25}{540}$.
$t = \frac{10 \times 540}{25} = \frac{5400}{25} = 216\, \text{s}$.

(A) According to $Newton's$ law of cooling, the rate of cooling is given by: $\frac{\theta_1 - \theta_2}{t} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right)$.
For the first interval: $\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 25 \right) \implies 1 = K(55 - 25) \implies 1 = 30K \implies K = \frac{1}{30} \dots (i)$.
For the second interval, let the final temperature be $\theta$: $\frac{50 - \theta}{10} = K \left( \frac{50 + \theta}{2} - 25 \right)$.
Substituting $K = \frac{1}{30}$: $\frac{50 - \theta}{10} = \frac{1}{30} \left( \frac{50 + \theta - 50}{2} \right) = \frac{1}{30} \left( \frac{\theta}{2} \right) = \frac{\theta}{60}$.
$6(50 - \theta) = \theta \implies 300 - 6\theta = \theta \implies 7\theta = 300 \implies \theta \approx 42.85 \, ^\circ\text{C} \approx 43 \, ^\circ\text{C}$.

(B) According to Newton's law of cooling,the time $t$ taken for a body to cool from temperature $\theta_1$ to $\theta_2$ in an environment with ambient temperature $\theta_0$ is given by $t = \frac{1}{k} \ln \left( \frac{\theta_1 - \theta_0}{\theta_2 - \theta_0} \right)$.
For the first interval ($80\,^oC$ to $40\,^oC$):
$5 = \frac{1}{k} \ln \left( \frac{80 - 30}{40 - 30} \right) = \frac{1}{k} \ln \left( \frac{50}{10} \right) = \frac{1}{k} \ln 5$.
So,$k = \frac{\ln 5}{5} = \frac{1.609}{5} = 0.3218\,min^{-1}$.
For the second interval ($62\,^oC$ to $32\,^oC$):
$t = \frac{1}{k} \ln \left( \frac{62 - 30}{32 - 30} \right) = \frac{1}{k} \ln \left( \frac{32}{2} \right) = \frac{1}{k} \ln 16 = \frac{1}{k} \ln(2^4) = \frac{4 \ln 2}{k}$.
Substituting the values:
$t = \frac{4 \times 0.693}{0.3218} \approx \frac{2.772}{0.3218} \approx 8.61\,minutes$.
Thus,the time taken is $8.6\,minutes$.

(B) According to $Newton's$ law of cooling,the rate of cooling is given by: $\frac{\theta_1 - \theta_2}{t} = K \left[ \frac{\theta_1 + \theta_2}{2} - \theta_0 \right]$,where $\theta_0$ is the surrounding temperature.
For the first $10$ minutes: $\frac{60 - 50}{10} = K \left[ \frac{60 + 50}{2} - \theta_0 \right] \implies 1 = K(55 - \theta_0) \dots (i)$
For the next $10$ minutes: $\frac{50 - 42}{10} = K \left[ \frac{50 + 42}{2} - \theta_0 \right] \implies 0.8 = K(46 - \theta_0) \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1}{0.8} = \frac{55 - \theta_0}{46 - \theta_0}$
$1.25 = \frac{55 - \theta_0}{46 - \theta_0}$
$1.25(46 - \theta_0) = 55 - \theta_0$
$57.5 - 1.25\theta_0 = 55 - \theta_0$
$2.5 = 0.25\theta_0$
$\theta_0 = 10\,^oC$.

(D) According to Newton's law of cooling,the rate of heat loss $\frac{dQ}{dt}$ depends on the temperature difference between the body and the surroundings. Since the vessel,surroundings,and temperature range are identical,the rate of heat loss is the same for both cases.
Let $m_w = 50\,g$ be the mass of water,$C_w = 1\,cal/(g \cdot ^oC)$ be the specific heat of water,$m_l = 100\,g$ be the mass of the liquid,$C_l$ be the specific heat of the liquid,and $W = 30\,g$ be the water equivalent of the vessel.
The total heat lost is given by $Q = (mC + W) \Delta T$.
Since the time $t$ is the same,the rate of heat loss is:
$\frac{(m_w C_w + W) \Delta T}{t} = \frac{(m_l C_l + W) \Delta T}{t}$
Canceling $\frac{\Delta T}{t}$ from both sides:
$m_w C_w + W = m_l C_l + W$
$m_w C_w = m_l C_l$
Substituting the values:
$50 \times 1 = 100 \times C_l$
$C_l = \frac{50}{100} = 0.5\,cal/(g \cdot ^oC) = 0.5\,kcal/(kg \cdot ^oC)$.