Newton's Law of Cooling Questions in English

(C) In forced convection,the rate of loss of heat $\frac{dQ}{dt}$ is governed by Newton's law of cooling,which states that the rate of heat loss is proportional to the surface area $A$ and the temperature difference between the body and its surroundings $(T - T_0)$.
Mathematically,$\frac{dQ}{dt} \propto A(T - T_0)$.
Therefore,the rate of loss of heat is proportional to the surface area $(B)$ and the excess of temperature over that of the surroundings $(D)$.
Thus,option $C$ is correct.

(D) According to Stefan-Boltzmann Law,the rate of heat loss $dQ/dt$ is proportional to the surface area $A$ for a given temperature $T$ and environment temperature $T_0$,i.e.,$dQ/dt \propto A(T^4 - T_0^4)$.
Since the rate of cooling $R = (dQ/dt) / (ms)$,where $m$ is mass and $s$ is specific heat,and both are the same for the sphere and the disc,the ratio of the rate of cooling is equal to the ratio of their surface areas.
Surface area of a sphere $A_s = 4\pi r^2$.
Surface area of a disc (considering both sides) $A_d = 2 \times (\pi r^2) = 2\pi r^2$.
Therefore,the ratio of the rate of cooling is $\frac{R_s}{R_d} = \frac{A_s}{A_d} = \frac{4\pi r^2}{2\pi r^2} = \frac{2}{1}$.

(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ is proportional to $(T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.
So,$\frac{R_1}{R_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.
Given $T_1 = 600 \ K$,$T_2 = 900 \ K$,$T_0 = 300 \ K$,and $R_1 = R$.
Substituting the values:
$\frac{R}{R_2} = \frac{(600)^4 - (300)^4}{(900)^4 - (300)^4} = \frac{(300)^4 (2^4 - 1^4)}{(300)^4 (3^4 - 1^4)} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16}$.
Therefore,$R_2 = \frac{16}{3}R$.

(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.
In the first case,the water cools from $60^oC$ to $50^oC$ in $10$ minutes:
$\frac{60 - 50}{10} = K \left[ \frac{60 + 50}{2} - \theta_0 \right]$
$1 = K(55 - \theta_0)$ --- $(i)$
In the second case,the water cools from $50^oC$ to $42^oC$ in the next $10$ minutes:
$\frac{50 - 42}{10} = K \left[ \frac{50 + 42}{2} - \theta_0 \right]$
$0.8 = K(46 - \theta_0)$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{1}{0.8} = \frac{55 - \theta_0}{46 - \theta_0}$
$1.25 = \frac{55 - \theta_0}{46 - \theta_0}$
$1.25(46 - \theta_0) = 55 - \theta_0$
$57.5 - 1.25\theta_0 = 55 - \theta_0$
$2.5 = 0.25\theta_0$
$\theta_0 = 10^oC$.

(C) According to Newton's law of cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundings.
Mathematically,$\frac{d\theta}{dt} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right)$,where $\theta_0$ is the room temperature.
Since the temperature difference decreases as the water cools,the rate of cooling decreases.
Therefore,the time taken to cool by the same amount $(5^{\circ} C)$ increases as the temperature of the water approaches the room temperature.
Comparing the mean temperatures: $\left( \frac{80+75}{2} \right) > \left( \frac{75+70}{2} \right) > \left( \frac{70+65}{2} \right)$.
Thus,the rate of cooling is highest for the first interval and lowest for the third interval.
Consequently,$t_1 < t_2 < t_3$.

(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,i.e.,$R \propto (\theta - \theta_0)$.
Given:
Temperature of body $B_1$,$\theta_1 = 100^oC$
Temperature of body $B_2$,$\theta_2 = 80^oC$
Temperature of surroundings,$\theta_0 = 40^oC$
At $t = 0$,the rates of cooling $R_1$ and $R_2$ are:
$R_1 \propto (\theta_1 - \theta_0) = (100 - 40) = 60$
$R_2 \propto (\theta_2 - \theta_0) = (80 - 40) = 40$
Therefore,the ratio is:
$\frac{R_1}{R_2} = \frac{60}{40} = \frac{3}{2}$
Thus,$R_1:R_2 = 3:2$.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = \frac{kA}{ms}(\theta - \theta_0)$.
Since the masses $m$ are equal and the calorimeters are similar (same surface area $A$ and emissivity $k$),the rate of cooling depends on the specific heat capacity $s$ of the liquids.
Specifically,$\frac{d\theta}{dt} \propto \frac{1}{s}$.
Therefore,the rate of cooling depends on the specific heats of the liquids.

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T - T_0)$.
For a calorimeter,the rate of heat loss is $\frac{dQ}{dt} = (mC + W) \frac{dT}{dt}$,where $m$ is the mass,$C$ is the specific heat,and $W$ is the water equivalent.
Since the cooling range $(70^{\circ}C$ to $60^{\circ}C)$ and the surrounding temperature are the same,the rate of heat loss $\frac{dQ}{dt}$ is constant for both.
Thus,$\frac{(m_W C_W + W) \Delta T}{t_W} = \frac{(m_L C_L + W) \Delta T}{t_L}$.
Rearranging for $C_L$: $C_L = \frac{1}{m_L} \left[ \frac{t_L}{t_W} (m_W C_W + W) - W \right]$.
Given: $m_W = 350 \ g$,$C_W = 1 \ Cal/g^{\circ}C$,$W = 10 \ g$,$t_W = 3 \ min = 180 \ s$,$m_L = 300 \ g$,$t_L = 95 \ s$.
Substituting the values: $C_L = \frac{1}{300} \left[ \frac{95}{180} (350 \times 1 + 10) - 10 \right]$.
$C_L = \frac{1}{300} \left[ \frac{95}{180} (360) - 10 \right] = \frac{1}{300} [95 \times 2 - 10] = \frac{190 - 10}{300} = \frac{180}{300} = 0.6 \ Cal/g^{\circ}C$.

(C) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = -k(T - T_s)$.
For the first interval: $\frac{60 - 50}{10} = k \left( \frac{60 + 50}{2} - 30 \right) \Rightarrow 1 = k(55 - 30) \Rightarrow 1 = 25k \Rightarrow k = \frac{1}{25}$.
For the second interval,let the final temperature be $T_f$: $\frac{50 - T_f}{10} = k \left( \frac{50 + T_f}{2} - 30 \right)$.
Substituting $k = \frac{1}{25}$: $\frac{50 - T_f}{10} = \frac{1}{25} \left( \frac{50 + T_f - 60}{2} \right) \Rightarrow \frac{50 - T_f}{10} = \frac{T_f - 10}{50}$.
Multiplying by $50$: $5(50 - T_f) = T_f - 10 \Rightarrow 250 - 5T_f = T_f - 10 \Rightarrow 6T_f = 260 \Rightarrow T_f = 43.33^{\circ}C$.
Since $43.33^{\circ}C > 40^{\circ}C$,the correct option is $C$.

(D) According to Newton's Law of Cooling,the rate of loss of heat $\left( \frac{dQ}{dt} \right)$ is directly proportional to the temperature difference $(\Delta \theta)$ between the liquid and its surroundings,provided the temperature difference is small.
Let $\left( \frac{dQ}{dt} \right)_1 = 60 \; cal/sec$ at $T_1 = 80^{\circ}C$.
The room temperature $T_0 = 20^{\circ}C$.
The temperature difference $\Delta \theta_1 = T_1 - T_0 = 80 - 20 = 60^{\circ}C$.
Let $\left( \frac{dQ}{dt} \right)_2$ be the rate of loss of heat at $T_2 = 40^{\circ}C$.
The temperature difference $\Delta \theta_2 = T_2 - T_0 = 40 - 20 = 20^{\circ}C$.
Using the proportionality $\frac{(dQ/dt)_1}{(dQ/dt)_2} = \frac{\Delta \theta_1}{\Delta \theta_2}$:
$\frac{60}{(dQ/dt)_2} = \frac{60}{20}$
$\frac{60}{(dQ/dt)_2} = 3$
$(dQ/dt)_2 = \frac{60}{3} = 20 \; cal/sec$.

(B) 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 interval: $\frac{100 - 70}{4} = K \left( \frac{100 + 70}{2} - 15 \right)$
$\frac{30}{4} = K (85 - 15) = 70K$
$K = \frac{30}{4 \times 70} = \frac{3}{28}$
For the second interval: $\frac{70 - 40}{t} = K \left( \frac{70 + 40}{2} - 15 \right)$
$\frac{30}{t} = \frac{3}{28} (55 - 15) = \frac{3}{28} \times 40$
$\frac{30}{t} = \frac{3 \times 10}{7} = \frac{30}{7}$
$t = 7 \text{ min}$

(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{ambient})$.
For the first interval: $\frac{80 - 60}{1} = K\left( \frac{80 + 60}{2} - 30 \right)$.
$20 = K(70 - 30) = 40K$,which gives $K = \frac{20}{40} = 0.5 \text{ min}^{-1}$.
For the second interval,let the time taken be $t$ minutes:
$\frac{60 - 50}{t} = 0.5 \left( \frac{60 + 50}{2} - 30 \right)$.
$\frac{10}{t} = 0.5(55 - 30) = 0.5 \times 25 = 12.5$.
$t = \frac{10}{12.5} = 0.8 \text{ minutes}$.
Converting to seconds: $t = 0.8 \times 60 = 48 \text{ sec}$.

(C) According to Newton's law of cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundings: $\frac{d\theta}{dt} = k(\theta - \theta_0)$.
In the first case,the liquid cools from $70^{\circ}C$ to $60^{\circ}C$. The average temperature is $\frac{70+60}{2} = 65^{\circ}C$. The temperature difference is $(65 - \theta_0)$.
In the second case,the liquid cools from $60^{\circ}C$ to $50^{\circ}C$. The average temperature is $\frac{60+50}{2} = 55^{\circ}C$. The temperature difference is $(55 - \theta_0)$.
Since the average temperature in the second case is lower,the temperature difference between the liquid and the surroundings is smaller. Consequently,the rate of cooling decreases. Therefore,it will take more time to cool through the same temperature range of $10^{\circ}C$.

(C) 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 $10 \; \text{minutes}$:
$\frac{62 - 50}{10} = K \left[ \frac{62 + 50}{2} - \theta_0 \right]$
$1.2 = K [56 - \theta_0] \quad \dots (i)$
For the next $10 \; \text{minutes}$:
$\frac{50 - 42}{10} = K \left[ \frac{50 + 42}{2} - \theta_0 \right]$
$0.8 = K [46 - \theta_0] \quad \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{1.2}{0.8} = \frac{56 - \theta_0}{46 - \theta_0}$
$1.5 = \frac{56 - \theta_0}{46 - \theta_0}$
$1.5(46 - \theta_0) = 56 - \theta_0$
$69 - 1.5\theta_0 = 56 - \theta_0$
$69 - 56 = 1.5\theta_0 - \theta_0$
$13 = 0.5\theta_0$
$\theta_0 = 26^{\circ}C$

(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = \frac{\sigma A}{mc}(T^4 - T_0^4)$.
Since the calorimeters and surroundings are identical,$A$ and $T_0$ are constant. Thus,$\frac{d\theta}{dt} \propto \frac{1}{mc}$.
Substituting mass $m = V\rho$,where $V$ is volume and $\rho$ is density,we get $\frac{d\theta}{dt} \propto \frac{1}{V\rho c}$.
For the rate of cooling to be the same for two different liquids,the product of density and specific heat capacity $(\rho c)$ must be equal,or the volumes must be adjusted such that $V_1 \rho_1 c_1 = V_2 \rho_2 c_2$.
Given the options,if we take equal volumes $(V_1 = V_2)$,the condition depends on the thermal properties. In many standard physics problems of this type,taking equal volumes is the intended condition to compare the cooling behavior based on their thermal capacities per unit volume. Thus,option $(d)$ is the correct choice.

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.
For the first $10$ minutes: $\frac{60 - 50}{10} = K\left( \frac{60 + 50}{2} - 25 \right) \Rightarrow 1 = K(55 - 25) \Rightarrow 1 = 30K$ --- $(i)$
For the next $10$ minutes,let the final temperature be $\theta$: $\frac{50 - \theta}{10} = K\left( \frac{50 + \theta}{2} - 25 \right) \Rightarrow \frac{50 - \theta}{10} = K\left( \frac{50 + \theta - 50}{2} \right) \Rightarrow \frac{50 - \theta}{10} = K\left( \frac{\theta}{2} \right)$ --- $(ii)$
Dividing $(i)$ by $(ii)$: $\frac{1}{(50 - \theta)/10} = \frac{30K}{K(\theta/2)} \Rightarrow \frac{10}{50 - \theta} = \frac{60}{\theta}$.
Cross-multiplying: $10\theta = 60(50 - \theta) \Rightarrow 10\theta = 3000 - 60\theta \Rightarrow 70\theta = 3000$.
$\theta = \frac{300}{7} \approx 42.85^{\circ}C$.

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_{room})$.
For the first interval: $\frac{365 - 361}{2} = K \left( \frac{365 + 361}{2} - 293 \right)$.
$2 = K(363 - 293) = K(70)$.
$K = \frac{2}{70} = \frac{1}{35} \ \text{min}^{-1}$.
For the second interval: $\frac{344 - 342}{t} = K \left( \frac{344 + 342}{2} - 293 \right)$.
$\frac{2}{t} = \frac{1}{35} (343 - 293) = \frac{1}{35} (50) = \frac{10}{7}$.
$t = \frac{2 \times 7}{10} = 1.4 \ \text{minutes}$.
$t = 1.4 \times 60 = 84 \ \text{seconds}$.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature.
For the first interval: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}{2} - 30.0 \right) \implies \frac{0.1}{5} = K(49.95 - 30.0) = K(19.95)$ ... $(i)$
For the second interval: $\frac{40.0 - 39.9}{t} = K \left( \frac{40.0 + 39.9}{2} - 30.0 \right) \implies \frac{0.1}{t} = K(39.95 - 30.0) = K(9.95)$ ... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{0.1/5}{0.1/t} = \frac{K(19.95)}{K(9.95)} \implies \frac{t}{5} = \frac{19.95}{9.95} \approx 2$
$t = 5 \times 2 = 10\;s$.

(C) 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 and $T_s$ is the temperature of the surroundings.
This implies that the time taken to cool down depends on the temperature difference between the body and the surroundings.
For the first interval,the average temperature is $\frac{100+80}{2} = 90^{\circ}C$.
For the second interval,the average temperature is $\frac{80+60}{2} = 70^{\circ}C$.
Since the temperature difference between the water and the surroundings is greater in the first interval,the rate of cooling is faster.
Therefore,the time taken to cool down by $20^{\circ}C$ is less in the first interval than in the second interval.
Thus,$T_1 < T_2$.

(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{-d\theta}{dt} = k \left( \theta_{avg} - \theta_0 \right)$,where $\theta_{avg} = \frac{\theta_1 + \theta_2}{2}$ is the average temperature of the body and $\theta_0$ is the surrounding temperature.
In the first case,the average temperature is $\frac{70 + 60}{2} = 65^{\circ}C$.
In the second case,the average temperature is $\frac{69 + 59}{2} = 64^{\circ}C$.
Since the average temperature in the second case is lower,the temperature difference between the body and the surroundings is smaller.
Therefore,the rate of cooling is slower in the second case.
Since the rate of cooling is lower,it will take more time to cool by the same $10^{\circ}C$ interval.
Thus,the time taken will be more than $4$ minutes.

(A) Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings.
This law is an empirical approximation that is valid only when the temperature difference between the body and the surroundings is small (typically less than $10^{\circ}C$ to $15^{\circ}C$).
When the temperature difference is small,the heat loss is primarily due to convection,which follows the linear relationship defined by the law.
Therefore,the correct option is $A$.

(B) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T - T_s)$.
For the first case: $\frac{61 - 59}{4} = K \left( \frac{61 + 59}{2} - 30 \right)$.
$\frac{2}{4} = K(60 - 30) \implies 0.5 = 30K \implies K = \frac{0.5}{30} = \frac{1}{60}$.
For the second case: $\frac{51 - 49}{t} = K \left( \frac{51 + 49}{2} - 30 \right)$.
$\frac{2}{t} = K(50 - 30) \implies \frac{2}{t} = 20K$.
Substituting $K = \frac{1}{60}$: $\frac{2}{t} = 20 \times \frac{1}{60} = \frac{1}{3}$.
Therefore,$t = 6 \text{ min}$.

(D) $Newton's$ $Law$ $of$ $cooling$ states that the rate of loss of heat $(dQ/dt)$ of a body is directly proportional to the difference in temperature between the body $(T)$ and its surroundings $(T_s)$,provided the temperature difference is small.
Mathematically,this is expressed as: $\frac{dQ}{dt} \propto (T - T_s)$.
Since the rate of cooling is defined as the rate of change of temperature $(dT/dt)$,it follows that $\frac{dT}{dt} \propto (T - T_s)$.
Therefore,the rate of cooling is proportional to the difference between the temperature of the body and the surroundings.

(C) According to Newton's Law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature.
For the first interval:
$\frac{60 - 40}{7} = K \left( \frac{60 + 40}{2} - 10 \right)$
$\frac{20}{7} = K(50 - 10) = 40K$
$K = \frac{20}{7 \times 40} = \frac{1}{14} \dots (i)$
For the second interval:
$\frac{40 - 28}{t} = K \left( \frac{40 + 28}{2} - 10 \right)$
$\frac{12}{t} = K(34 - 10) = 24K$
$K = \frac{12}{24t} = \frac{1}{2t} \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{1}{14} = \frac{1}{2t}$
$2t = 14$
$t = 7 \text{ minutes}$.

(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: $\frac{dT}{dt} = K(T_{avg} - T_s)$.
For the first interval: $\frac{50 - 40}{5} = K \left( \frac{50 + 40}{2} - T_s \right) \implies 2 = K(45 - T_s)$ .... $(i)$
For the second interval: $\frac{40 - 33.33}{5} = K \left( \frac{40 + 33.33}{2} - T_s \right) \implies \frac{6.67}{5} = K(36.665 - T_s) \implies 1.334 = K(36.665 - T_s)$ .... $(ii)$
Dividing equation $(i)$ by $(ii)$: $\frac{2}{1.334} = \frac{45 - T_s}{36.665 - T_s}$.
$1.5 \approx \frac{45 - T_s}{36.665 - T_s} \implies 1.5(36.665 - T_s) = 45 - T_s$.
$54.9975 - 1.5T_s = 45 - T_s \implies 0.5T_s = 9.9975 \implies T_s \approx 20^oC$.

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and the surroundings: $\frac{dT}{dt} = K(T_{avg} - T_s)$.
In the first $10$ minutes,the temperature falls from $50^oC$ to $40^oC$ with surroundings at $20^oC$:
$\frac{50 - 40}{10} = K \left( \frac{50 + 40}{2} - 20 \right) \implies 1 = K(45 - 20) \implies 1 = 25K \implies K = \frac{1}{25} \dots (i)$
In the next $10$ minutes,let the final temperature be $\theta_2$:
$\frac{40 - \theta_2}{10} = K \left( \frac{40 + \theta_2}{2} - 20 \right) \dots (ii)$
Substituting $K = \frac{1}{25}$ into equation $(ii)$:
$\frac{40 - \theta_2}{10} = \frac{1}{25} \left( 20 + \frac{\theta_2}{2} - 20 \right) \implies \frac{40 - \theta_2}{10} = \frac{\theta_2}{50}$
$5(40 - \theta_2) = \theta_2 \implies 200 - 5\theta_2 = \theta_2 \implies 6\theta_2 = 200$
$\theta_2 = \frac{200}{6} = 33.33^oC$.

(A) According to Newton's law of cooling,the rate of loss of heat (or rate of cooling) of a body is directly proportional to the temperature difference between the body and its surroundings for small temperature differences.
Mathematically,this is expressed as: $\frac{dQ}{dt} \propto \Delta \theta$.
Comparing this with the given expression $(\Delta \theta)^n$,we find that $n = 1$.

(B) According to Newton's law of cooling,the rate of change of temperature 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 interval ($t = 5$ minutes):
$\frac{80 - 64}{5} = K \left[ \frac{80 + 64}{2} - \theta_0 \right] \implies 3.2 = K(72 - \theta_0)$ ... $(i)$
For the second interval ($t = 10$ minutes,from $64^{\circ}C$ to $52^{\circ}C$):
$\frac{64 - 52}{10} = K \left[ \frac{64 + 52}{2} - \theta_0 \right] \implies 1.2 = K(58 - \theta_0)$ ... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{3.2}{1.2} = \frac{72 - \theta_0}{58 - \theta_0} \implies \frac{8}{3} = \frac{72 - \theta_0}{58 - \theta_0}$
$8(58 - \theta_0) = 3(72 - \theta_0) \implies 464 - 8\theta_0 = 216 - 3\theta_0$
$5\theta_0 = 248 \implies \theta_0 = 49.6^{\circ}C$. Given the options,the closest value is $49^{\circ}C$.

(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T_{avg} - T_0)$.
For the first interval: $\frac{50 - 45}{5} = K(\frac{50 + 45}{2} - T_0) \implies 1 = K(47.5 - T_0)$ ... $(i)$
For the second interval: $\frac{45 - 41.5}{5} = K(\frac{45 + 41.5}{2} - T_0) \implies 0.7 = K(43.25 - T_0)$ ... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{1}{0.7} = \frac{47.5 - T_0}{43.25 - T_0}$
$43.25 - T_0 = 0.7(47.5 - T_0)$
$43.25 - T_0 = 33.25 - 0.7T_0$
$0.3T_0 = 10$
$T_0 = \frac{10}{0.3} = 33.33\;^oC$.
Thus,the surrounding temperature is $33.3\;^oC$.

(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.
For the first case: $\frac{65.5 - 62.5}{1} = K \left( \frac{65.5 + 62.5}{2} - 22.5 \right)$.
$3 = K(64 - 22.5) = K(41.5)$,so $K = \frac{3}{41.5}$.
For the second case: $\frac{46.5 - 40.5}{t} = K \left( \frac{46.5 + 40.5}{2} - 22.5 \right)$.
$\frac{6}{t} = \frac{3}{41.5} (43.5 - 22.5) = \frac{3}{41.5} (21)$.
$t = \frac{6 \times 41.5}{3 \times 21} = \frac{2 \times 41.5}{21} = \frac{83}{21} \approx 3.95 \approx 4$ minutes.

(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$.
For the first $10$ minutes: $\frac{62 - 50}{10} = K\left( \frac{62 + 50}{2} - 26 \right)$.
$\frac{12}{10} = K(56 - 26) \implies 1.2 = K(30) \implies K = \frac{1.2}{30} = 0.04$.
For the next $10$ minutes,let the final temperature be $\theta$:
$\frac{50 - \theta}{10} = 0.04 \left( \frac{50 + \theta}{2} - 26 \right)$.
$50 - \theta = 0.4 \left( 25 + 0.5\theta - 26 \right)$.
$50 - \theta = 0.4(0.5\theta - 1) = 0.2\theta - 0.4$.
$50.4 = 1.2\theta \implies \theta = \frac{50.4}{1.2} = 42^\circ C$.

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

(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surrounding})$.
In the first case,the temperature changes from $75^{\circ}C$ to $65^{\circ}C$ in $t_1 = 2$ minutes.
$\frac{75 - 65}{2} = K \left( \frac{75 + 65}{2} - 30 \right)$
$\frac{10}{2} = K(70 - 30) \implies 5 = 40K \implies K = \frac{5}{40} = \frac{1}{8}$ $(i)$
In the second case,the temperature changes from $55^{\circ}C$ to $45^{\circ}C$ in $t_2$ minutes.
$\frac{55 - 45}{t_2} = K \left( \frac{55 + 45}{2} - 30 \right)$
$\frac{10}{t_2} = K(50 - 30) \implies \frac{10}{t_2} = 20K$ (ii)
Dividing equation $(i)$ by (ii):
$\frac{5}{10/t_2} = \frac{40K}{20K}$
$\frac{5t_2}{10} = 2$
$5t_2 = 20 \implies t_2 = 4$ minutes.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.
In the first case,the body cools from $80^{\circ}C$ to $50^{\circ}C$ in $5 \text{ min}$.
$\frac{80 - 50}{5} = K \left( \frac{80 + 50}{2} - 20 \right)$
$\frac{30}{5} = K(65 - 20) \implies 6 = 45K \implies K = \frac{6}{45} = \frac{2}{15} \dots (i)$
In the second case,the body cools from $60^{\circ}C$ to $30^{\circ}C$ in $t \text{ min}$.
$\frac{60 - 30}{t} = K \left( \frac{60 + 30}{2} - 20 \right)$
$\frac{30}{t} = K(45 - 20) \implies \frac{30}{t} = 25K \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{6}{30/t} = \frac{45K}{25K}$
$\frac{6t}{30} = \frac{45}{25}$
$\frac{t}{5} = \frac{9}{5}$
$t = 9 \text{ min}$.

(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: $\frac{dT}{dt} = -k(T - T_s)$.
This implies that the rate of heating (or cooling) is higher when the temperature difference between the object and the surroundings is larger.
In the first case,the temperature range is $0^{\circ}C$ to $5^{\circ}C$. The average temperature is $2.5^{\circ}C$,so the average temperature difference is $25^{\circ}C - 2.5^{\circ}C = 22.5^{\circ}C$.
In the second case,the temperature range is $10^{\circ}C$ to $15^{\circ}C$. The average temperature is $12.5^{\circ}C$,so the average temperature difference is $25^{\circ}C - 12.5^{\circ}C = 12.5^{\circ}C$.
Since the temperature difference is larger in the first case,the rate of heat transfer is higher,meaning the time taken $t_1$ is less than the time taken $t_2$ $(t_1 < t_2)$.

(B) The rate of cooling $R$ of a body is given by $R = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.
Since the material is the same,the emissivity $\varepsilon$,specific heat $c$,and density $\rho$ are the same.
Thus,$R \propto \frac{A}{m} \propto \frac{A}{\rho V} \propto \frac{A}{V}$,where $A$ is the surface area and $V$ is the volume.
For a given surface area $A$,the rate of cooling $R \propto \frac{1}{V}$.
For a cube of side $a$,$A = 6a^2$ and $V = a^3 = (\sqrt{A/6})^3$.
For a sphere of radius $r$,$A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\sqrt{A/4\pi})^3$.
Comparing volumes for the same surface area,the volume of the cube is less than the volume of the sphere $(V_{cube} < V_{sphere})$.
Therefore,the rate of cooling of the cube is greater than that of the sphere $(R_{cube} > R_{sphere})$.

(B) The rate of cooling $R$ is given by the formula $R = \frac{d\theta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.
Since both spheres are made of the same material and have the same surface finish,$\epsilon$,$\sigma$,$T$,$T_0$,and the density $\rho$ are constant.
Thus,$R \propto \frac{A}{m}$.
Since $A = 4\pi r^2$ and $m = \rho \cdot \frac{4}{3}\pi r^3$,we have $A \propto r^2$ and $m \propto r^3$,which implies $r \propto m^{1/3}$.
Substituting this into the proportionality,$R \propto \frac{r^2}{r^3} \propto \frac{1}{r} \propto \frac{1}{m^{1/3}}$.
Therefore,the ratio of the rates of cooling is $\frac{R_1}{R_2} = \left(\frac{m_2}{m_1}\right)^{1/3}$.
Given $m_1 = 3m_2$,we get $\frac{R_1}{R_2} = \left(\frac{m_2}{3m_2}\right)^{1/3} = \left(\frac{1}{3}\right)^{1/3}$.

(A) The rate of cooling is given by the formula: $\frac{d\theta}{dt} = \frac{P}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.
Since both spheres have the same material,size (surface area $A$),and are in the same surroundings,the rate of heat loss $P$ is the same for both.
The rate of cooling $\frac{d\theta}{dt}$ is inversely proportional to the mass $m$ of the object: $\frac{d\theta}{dt} \propto \frac{1}{m}$.
Since the solid sphere has a larger mass $(m_{solid} > m_{hollow})$,its rate of cooling will be lower.
Therefore,the hollow sphere,having a smaller mass,will cool at a faster rate for all values of $T$.

(C) The rate of cooling is given by $\frac{d\theta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.
Since the temperature change $\Delta \theta$ is small,we can write $\frac{\Delta \theta}{t} \propto \frac{A}{mc}$.
Here,$m = \rho V = \rho a^3$ and $A = 6a^2$,where $a$ is the edge length.
Thus,$t \propto \frac{mc}{A} \propto \frac{\rho a^3 c}{6a^2} \propto a$.
Therefore,$\frac{t_1}{t_2} = \frac{a_1}{a_2}$.
Given $t_1 = 100\;s$,$a_1 = 1\;cm$,and $a_2 = 2\;cm$.
$\frac{100}{t_2} = \frac{1}{2} \implies t_2 = 200\;s$.

(A) 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 interval ($0$ to $5$ min): $\frac{80 - 64}{5} = K \left[ \frac{80 + 64}{2} - \theta_0 \right] \implies 3.2 = K(72 - \theta_0)$ ... $(i)$
For the second interval ($0$ to $10$ min): $\frac{80 - 52}{10} = K \left[ \frac{80 + 52}{2} - \theta_0 \right] \implies 2.8 = K(66 - \theta_0)$ ... $(ii)$
Dividing $(i)$ by $(ii)$: $\frac{3.2}{2.8} = \frac{72 - \theta_0}{66 - \theta_0} \implies \frac{8}{7} = \frac{72 - \theta_0}{66 - \theta_0}$.
$8(66 - \theta_0) = 7(72 - \theta_0) \implies 528 - 8\theta_0 = 504 - 7\theta_0 \implies \theta_0 = 24^\circ C$.
Substituting $\theta_0 = 24$ in $(i)$: $3.2 = K(72 - 24) \implies 3.2 = K(48) \implies K = \frac{3.2}{48} = \frac{1}{15}$.
For the third interval ($0$ to $15$ min): $\frac{80 - \theta}{15} = K \left[ \frac{80 + \theta}{2} - 24 \right]$.
$\frac{80 - \theta}{15} = \frac{1}{15} \left[ \frac{80 + \theta - 48}{2} \right] \implies 80 - \theta = \frac{32 + \theta}{2}$.
$160 - 2\theta = 32 + \theta \implies 3\theta = 128 \implies \theta = 42.66^\circ C \approx 42.7^\circ C$.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k$ is a positive constant.
In the $\theta-t$ graph,the slope of the tangent at any point is $\frac{d\theta}{dt}$. Since the curve is decreasing,the slope is negative. The angle $\varphi$ is the angle made by the tangent with the positive $t$-axis,so the slope is $-\tan \varphi$ (as the tangent makes an obtuse angle with the positive $t$-axis,or we consider the magnitude of the slope as $\tan \varphi$ where $\varphi$ is the angle with the negative $t$-axis direction).
At point $P$ where $\theta = \theta_2$,the magnitude of the slope is $\tan \varphi_2 = k(\theta_2 - \theta_0)$.
At point $Q$ where $\theta = \theta_1$,the magnitude of the slope is $\tan \varphi_1 = k(\theta_1 - \theta_0)$.
Dividing the two equations,we get: $\frac{\tan \varphi_2}{\tan \varphi_1} = \frac{k(\theta_2 - \theta_0)}{k(\theta_1 - \theta_0)} = \frac{\theta_2 - \theta_0}{\theta_1 - \theta_0}$.

(C) According to Newton's law of cooling,for a small temperature difference between a body and its surroundings,the rate of loss of heat $R$ is directly proportional to the temperature difference $(\theta - \theta_0)$.
$R \propto (\theta - \theta_0)$
$R = k(\theta - \theta_0) = k\theta - k\theta_0$,where $k$ is a positive constant.
Comparing this with the equation of a straight line $y = mx + c$,where $y = R$,$x = \theta$,$m = k$ (slope),and $c = -k\theta_0$ (y-intercept).
Since $k$ and $\theta_0$ are positive,the intercept $c = -k\theta_0$ is negative.
Therefore,the graph is a straight line with a positive slope and a negative y-intercept,which corresponds to the graph shown in option $C$.

(A) According to Newton's law of cooling,the rate of loss of heat is given by $\frac{dQ}{dt} = \varepsilon A \sigma (T^4 - T_0^4) \approx \varepsilon A \sigma (4T_0^3) \Delta T$.
Since the rate of loss of heat is also $\frac{dQ}{dt} = mc \frac{dT}{dt}$,we have $mc \frac{dT}{dt} = \varepsilon A \sigma (4T_0^3) \Delta T$.
Thus,the rate of fall of temperature is $\frac{dT}{dt} = \frac{\varepsilon A \sigma 4T_0^3}{mc} \Delta T$.
For both the sphere and the cube,the surface area $A$ is the same. However,the volume $V$ of a sphere is less than the volume of a cube for the same surface area. Since the mass $m = \rho V$,the mass of water in the sphere is less than the mass of water in the cube $(m_S < m_C)$.
Since $\frac{dT}{dt} \propto \frac{1}{m}$,the rate of cooling is faster for the sphere $(S)$ than for the cube $(C)$.
Therefore,the temperature of the sphere drops faster than the temperature of the cube,which is represented by the graph where the curve for $S$ is steeper than the curve for $C$.

(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,we get:
$\frac{d\theta}{\theta - \theta_0} = -k dt$
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 integration constant.
Therefore,the graph of $\log_e(\theta - \theta_0)$ versus $t$ is a straight line with a negative slope.

(D) Newton's law of cooling states that the rate of loss of heat $(dQ/dt)$ of a body is directly proportional to the difference in temperature between the body $(T)$ and its surroundings $(T_s)$,provided the difference is small.
Mathematically,$dQ/dt \propto (T - T_s)$.
Since the rate of cooling is defined as the rate of change of temperature $(dT/dt)$,it follows that $dT/dt \propto (T - T_s)$.
Therefore,the rate of cooling is directly proportional to the temperature difference between the body and the surroundings.

(C) 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 between the water and the surroundings decreases.
Consequently,the rate of cooling decreases,meaning the time taken to cool by a specific temperature interval increases.
Since the initial temperature is highest for the first interval ($75^{\circ}C$ to $70^{\circ}C$),the rate of cooling is fastest,resulting in the shortest time $T_1$.
As the water cools further,the temperature difference becomes smaller,leading to slower cooling and longer time intervals.
Therefore,the relationship between the times is $T_1 < T_2 < T_3$.

(C) Newton's law of cooling states that the rate of loss of heat $dQ/dt$ of a body is directly proportional to the temperature difference between the body and its surroundings,i.e.,$dQ/dt = -k(\theta - \theta_0)$.
Since the rate of loss of heat is also given by $dQ/dt = -ms(d\theta/dt)$,where $m$ is the mass,$s$ is the specific heat,and $d\theta/dt$ is the rate of cooling.
By equating these,we get $-ms(d\theta/dt) = -k(\theta - \theta_0)$.
This relationship allows us to determine the specific heat capacity '$s$' of a liquid if the other parameters are known.

(C) According to Newton's Law of Cooling,the rate of heat loss $(dQ/dt)$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$,where $e$ is emissivity,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,$T$ is the temperature of the body,and $T_0$ is the temperature of the surroundings.
Since both spheres are made of the same metal and are heated to the same temperature $T$ and placed in the same environment $T_0$,the rate of heat loss depends on the surface area $A$ and emissivity $e$.
If the spheres have the same radius,they have the same surface area $A = 4\pi r^2$.
Assuming the spheres have the same radius,the rate of heat loss will be the same for both,provided the emissivity is also the same.

(D) Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings,provided the difference is small.
This law is an empirical approximation derived from the Stefan-Boltzmann law,which describes the power radiated from a black body in terms of its temperature.
Specifically,when the temperature difference $\Delta T$ between the body and the surroundings is small,the Stefan-Boltzmann law can be linearized to show that the rate of cooling is proportional to $\Delta T$,which is the essence of Newton's law of cooling.

(B) According to Newton's law of cooling,the rate of heat loss is $\frac{dQ}{dt} = 4 \sigma A \theta_0^3 (\Delta \theta)$.
Since $\frac{dQ}{dt} = ms \left( -\frac{d\theta}{dt} \right)$,the rate of cooling is $-\frac{d\theta}{dt} = \frac{4 \sigma A \theta_0^3 (\Delta \theta)}{ms}$.
Here,mass $m = \text{Volume} \times \text{density} = (\frac{4}{3} \pi r^3) d$ and surface area $A = 4 \pi r^2$.
Substituting these values: $-\frac{d\theta}{dt} = \frac{4 \sigma (4 \pi r^2) \theta_0^3 (\Delta \theta)}{(\frac{4}{3} \pi r^3) d s}$.
Simplifying the expression: $-\frac{d\theta}{dt} = \frac{16 \pi r^2 \sigma \theta_0^3 (\Delta \theta)}{\frac{4}{3} \pi r^3 d s} = \frac{12 \sigma \theta_0^3 \Delta \theta}{rds}$.