NEET Physics Class 11 Chapter 8 Newton’s Law Of Cooling

NEET Physics Class 11 Chapter 8 Newton’s Law Of Cooling

For a small temperature difference between a body and its surroundings, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

⇒ \(\frac{d Q}{d t} \propto\left(\theta-\theta_0\right)\), where θ and θ0are temperature corresponding to object and surroundings.

From above expression \(\frac{d \theta}{d t}=-k\left(\theta-\theta_0\right)\)

This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law, which gives,

⇒ \(\mathrm{k}=\frac{4 \mathrm{e} \sigma \theta_0^3}{\mathrm{mc}} \mathrm{A}\)

Now \(\frac{d \theta}{d t}=-k\left[\begin{array}{ll}
\theta & \left.-\theta_0\right]
\end{array}\right.\)

⇒ \(\int_{\theta_i}^{\theta_f} \frac{d \theta}{\left(\theta-\theta_0\right)}=\int_0^t-k d t\)

where \(\theta_{\mathrm{i}}=\ln \frac{\left(\theta_f-\theta_0\right)}{\left(\theta_i-\theta_0\right)}\) initial temperature of object and

θf = final temperature of object.

⇒ –kt ⇒ (θf− θ0) = (θi– θ0) e–kt

⇒ θf= θ0+ (θi– θ0) e –kt

Limitations of Newton’s Law of Cooling:

1. The difference in temperature between the body and surroundings must be small
2. The loss of heat from the body should be by radiation only.
3. The temperature of the surroundings must remain constant during the cooling of the body.

Approximate Method For Applying Newton’s Law Of Cooling

Sometimes when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle-\mathrm{k}\left(<\theta>-\theta_0\right)\)

If θ_i and θ_f are the initial and final temperatures of the body then,

⇒ \(<\theta>=\frac{\theta_i+\theta_f}{2}\)

Remember equation is only an approximation and an equation must be used for exact values.

Comparison Of Specific Heat Of Two Liquids Using Newton’s Law Of Cooling:

If an equal volume of two liquids of densities and specific heats ρ₁, s_1, and ρ₂, s₂ respectively are filled in calorimeters having the same surface area and finish, cool from the same initial temperature θ₁ to the same final temperature θ₂ with the same temperature of surroundings, i.e., θ₀, in time intervals t₁and t₂ respectively and water equivalent of calorimeter is w. According to Newton’s law of cooling

⇒ \(\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liq }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}\)

⇒ \(\frac{\left(w+m_1 s_1\right)\left(\theta_1-\theta_2\right)}{t_1}\)

⇒ \(\frac{\left(w+m_2 s_2\right)\left(\theta_1-\theta_2\right)}{t_2}\)

or \(\frac{w+m_1 s_1}{t_1}=\frac{w+m_2 s_2}{t_2}\)

If the water equivalent of calorimeter w is negligible then

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

So, \(\frac{m_1 s_1}{m_2 s_2}=\frac{t_1}{t_2} \quad \text { or } \frac{\rho_1 s_1}{\rho_2 s_2}=\frac{t_1}{t_2}\)(v₁= v₂, volume are equal) with the help of this eqn. we can find the specific heat of the liquid.

Question 1. A body at a temperature of 40ºC is kept in a surrounding constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Answer:

Δθ_f = Δθ_ie^-kt

for the interval in which temperature falls from 40 to 35º C.

(35 – 20) = (40 – 20) e^-k.10

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{3}{4}\)

⇒ \(K=\frac{\ln \frac{4}{3}}{10}\)

for the next interval (30 – 20) = (35 – 20)e^-kt

⇒ \(\mathrm{e}^{-10 \mathrm{k}}=\frac{2}{3}\)

⇒ \(\mathrm{kt}=\mathrm{ln} \frac{3}{2}\)

⇒ \(\frac{\left(\ln \frac{4}{3}\right) t}{10}=\ell n\)

⇒ \(\mathrm{t}=10 \frac{\left(\ln \frac{3}{2}\right)}{\left(\ln \frac{4}{3}\right)}\) minute = 14.096 min

Alternative : (by approximate method) for the interval in which temperature falls from 40 to 35ºC 40 35

<θ> = \(\frac{40+35}{2}\) = 37.5ºC

⇒ \(\left\langle\frac{d \theta}{d t}\right\rangle=-\mathrm{k}\left(<\theta>-\theta_0\right)\)

⇒ \(\frac{\left(35^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}\right)}{10(\mathrm{~min})}\)

= –K(37.5ºC – 20ºC)

K = \(\frac{1}{35}\)(min^-1)

for the interval in which temperature falls from 35ºC to 30ºC

<θ> = \(\frac{\left(30^{\circ} \mathrm{C}-35^{\circ} \mathrm{C}\right)}{t}\) = (32.5º) C

= – K(32.5ºC – 20ºC)

⇒ required time, t = \(\)

⇒ \(\frac{5}{12.5} \times 35 \mathrm{~min}\)

= 14 min

Question 2. Two liquids of the same volume are cooled under the same conditions from 65ºC to 50ºC. The time taken is 200sec and 480sec. If the ratio of their specific heats is 2 : 3 then find the ratio of their densities. (neglect the water equivalent of a calorimeter)
Answer:

From Newton’s law of cooling

⇒ \(\left(\frac{m_1 s_1+w_1}{t_1}\right)\left(\theta_1-\theta_2\right)\)

⇒ \(\left(\frac{m_2 s_2+w_2}{t_2}\right)\left(\theta_1-\theta_2\right)\)

here w₁= w₂= 0

⇒ \(\frac{m_1 s_1}{t_1}=\frac{m_2 s_2}{t_2}\)

⇒ \(\frac{V d_1 s_1}{t_1}=\frac{V d_2 s_2}{t_2}\)

⇒ \(\frac{d_1}{d_2}=\frac{t_1 s_2}{t_2 s_1}\)

⇒ \(\frac{200}{480} \times \frac{3}{2}\)

⇒ \(\frac{5}{8}\)

Question 3. A calorimeter of water equivalent to 5 × 10^-3 kg contains 25 × 10^-3 kg of water. It takes 3 minutes to cool from 28°C to 21°C. When the same calorimeter is filled with 30 × 10^-3 kg of turpentine oil then it takes 2 minutes to cool from 28°C to 21°C. Find out the specific heat of turpentine oil.
Answer:

⇒ \(\mathrm{R}_{\text {water }}=\mathrm{R}_{\text {turpentine }} \frac{\left(m_1+w\right)}{t_1}=\frac{\left(m_2 s_2+w\right)}{t_2}\)

or \(\frac{\left(25 \times 10^{-3}+5 \times 10^{-3}\right)}{3}=\frac{30 \times 10^{-3} s_2+5 \times 10^{-3}}{2}\)

10 = \(\frac{30 s_2+5}{2}, 20\)

⇒ \(30 s_2+5\)

∴ specific heat of turpentine s₂= 1/2 = 0.5 kcal/kg/°C

Question 4. A man, the surface area of whose skin is 2m², is sitting in a room where the air temperature is 20°C. If his skin temperature is 28°C, find the rate at which his body loses heat. The emissivity of his skin = 0.97.
Answer:

Absolute room temperature (T₀) = 20 + 273 = 293 K

Absolute skin temperature (T) = 28 + 273 = 301 K

Rate of heat loss = σ e A (T⁴ – T₀⁴)

= 5.67 × 10^-8 × 0.97 × 2 × {(301)⁴ – (293)⁴} = 92.2 W

Question 5. Compare the rate of loss of heat from a metal sphere of the temperature 827°C, with the rate of loss of heat from the same sphere at 427 °C, if the temperature of surroundings is 27°C.
Answer:

Given : T₁= 827 °C = 1100 K, T₂= 427 °C = 700 K and T₀= 27 °C = 300 K

According to Steafan’s law of radiation, \(\frac{d Q}{d t}\)

⇒ \(\sigma \mathrm{Ae}\left(\mathrm{T}^4-\mathrm{T}_0{ }^4\right)\)

⇒ \(\frac{\left(\frac{d Q}{d t}\right)_1}{\left(\frac{d Q}{d t}\right)_2}=\frac{\left(T_1^4-T_0^4\right)}{\left(T_2^4-T_0^4\right)}\)

⇒ \(\frac{\left[(1100)^4-(300)^4\right]}{\left[(700)^4-(300)^4\right]}\) = 6.276

⇒ \(\left(\frac{d Q}{d t}\right)_1:\left(\frac{d Q}{d t}\right)_2\) = 6.276:1

Question 6. Two spheres of the same material have radii of 6 cm and 9 cm respectively. They are heated to the same temperature and allowed to cool in the same enclosure. Compare their initial rates of emission of heat and initial rates of fall of temperature.
Answer:

Given : r₁= 6 cm r₂= 9 cm,

∴ \(\frac{r_1}{r_2}=\frac{2}{3}\)

According to Stefan’s law of radiation, the rate of emission of heat by an ordinary body,

⇒ \(\mathrm{R}_{\mathrm{h}}=\left(\frac{d Q}{d t}\right)=\sigma \mathrm{AeT}^4\)

or \(R_h \propto r^2\) (A=4σr²)

⇒ \(\frac{R_{h 1}}{R_{h 2}}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

⇒ \(\frac{R_{F T 1}}{R_{F T 2}}=\frac{r_2}{r_1}=\frac{3}{2}\)

[Rate of fall of temp. \(\mathrm{R}_{\mathrm{FT}}, \frac{d \theta}{d t}, \frac{\sigma A e\left(T^4-T_0^4\right)}{m s J}\) = or \(R_{F T} \propto \frac{A}{m} \text { or } R_{F T} \propto \frac{1}{r}\)]

∴ Initial rates of emission of heat are in the ratio 4: 9 and initial rates of fall of temperature are in the ratio 3: 2.

Question 7. The filament of an evacuated light bulb has a length of 10 cm, a diameter of 0.2 mm, and an emissivity of 0.2, calculate the power it radiates at 2000 K. (σ = 5.67 × 10^-8 W/m² K⁴)
Answer:

l = 10 cm = 0.1 m, d = 0.2 mm, r = 0.1 mm = 1 × 10^-4 m, e = 0.2, T = 2000 K, σ = 5.67 × 10^-8 W/m² K⁴

According to Stefan’s law of radiation, the rate of emission of heat for an ordinary body (filament), E = σAeT⁴ = σ(2 π r l) eT⁴

= 5.67 × 10^-8 × 2 × 3.14 × 1 × 10^-4 × 0.1 × 0.2 × (2000)⁴

= 11.4 W

∴ Power radiated by the filament = 11.4 W [A = 2πrl]

Question 8. The energy radiated from a black body at a temperature of 727°C is E. By what factor does the radiated energy increase if the temperature is raised to 2227°C?
Solution :

⇒ \(\frac{E_2}{E_1}=\left[\frac{T_2}{T_1}\right]^4\)

⇒ \(\left[\frac{2227+273}{727+273}\right]^4=\left[\frac{2500}{1000}\right]^4\)

= 39

Question 9. An ice box made of 1.5 cm thick styrofoam has dimensions of 60 cm × 30cm. It contains ice at 0ºC and is kept in a room at 40ºC. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36 × 10⁵J/kg. and thermal conductivity of stryrofoam = 0.4 W/m-ºC.
Answer:

The total surface area of the walls

= 2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm)

= 1.44 m²

The thickness of the wall = 1.5 cm = 0.015m

The rate of heat flow into the box is

⇒ \(\frac{\Delta Q}{\Delta t}=\frac{K A\left(\theta_1-\theta_2\right)}{x}\)

⇒ \(\frac{\left(0.04 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(1.44 \mathrm{~m}^2\right)\left(40^{\circ} \mathrm{C}\right)}{0.015 \mathrm{~m}}\)

= 154 W.

The rate at which the ice melts is = 0.46 g/s

Question 10. A black body emits 10 watts per cm² at 327ºC. The sun radiates 10⁵ watts per cm². Then what is the temperature of the sun?
Answer:

∴ \(\frac{E_{\text {sun }}}{E_{\text {body }}}=\left(\frac{E_{\text {sun }}}{E_{\text {body }}}\right)^4\)

∴ \(\frac{T_{\text {sun }}}{T_{\text {body }}}=\left(\frac{10^5}{10}\right)^{1 / 4}\)

= 6000 K

∴ T_sun = 6000 K

Question 11. A bulb made of a tungsten filament of a surface is 0.5 cm² is heated to a temperature of 3000k when operated at 220V. The emissivity of the filament is e = 0.35 and take σ = 5.7 × 10^-8 mks units. Then the wattage of the bulb is …..(calculate)
Answer:

The emissive power watt/m² is

E = eσT⁴

Therefore the power of the bulb is

P = E x area (Watts)

∴ P = eσT⁴A

∴ P = 0.35 × 0.5 × 10^-4× 5, 7 × 10^-8× (3000)⁴

⇒ P = 80.8 W

Question 12. In the above example, if the temperature of the filament falls to 2000k due to a drop in mains voltage, then what will be the wattage of the bulb?
Answer:

Now the power of the bulb will be such that

⇒ \(\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^4\)

Thus \(P_2=P_1 \times\left(\frac{2}{3}\right)^4\)

∴ \(P_2=80.8 \times \frac{16}{81}\)

Thus P₂= P1x

∴ P₂= 80.8 x

⇒ P₂= 15.96

Question 13. A liquid takes 30 seconds to cool from 95ºC to 90ºC and 70 seconds to cool from 55 to 50ºC. Find the room temperature and the time it will take to cool from 50ºC to 45ºC
Answer:

From the first date

⇒ \(\frac{95-90}{30}=\mathrm{k}\left(\frac{95 \times 90}{2}-\theta_0\right)=\mathrm{k}\)….(1)

From the second data \(\frac{55-50}{70}=\mathrm{k}\left(\frac{55-50}{0}-\theta_0\right)\) = k….(2)

Dividing (1) and (2) we get

⇒ \(\frac{7}{3}=\frac{92.5-\theta_0}{52.5-\theta_0}\)

⇒ \(\theta_0=22.5^{\circ}\) ……….(3)

Let the time taken in cooling from 50ºC to 45ºC is t, then

⇒ \(\frac{50-45}{t}\)

⇒ \(\mathrm{k}\left(\frac{50-45}{2}-\theta_0\right)\)….(4)

Using θ₀= 22.5ºC, and dividing (1) by (2) we get

⇒ \(\frac{t}{30}=\left(\frac{92.5-22.5}{47.5-22.5}\right)\)

t = 84 sec

Question 14. A blackened metal disc is held normal to the sun’s rays, Both of its surfaces are exposed to the atmosphere if the distance of the earth from the sun is 216 times the radius of the sun and the temperature of the sun is 6000K, the temperature of the disc in the steady state will be
Answer:

In the steady state, the heat received from the sun will be equal to the heat radiated out. Heat received from the sun will be on one side only and it will radiate from both sides.

∴ \(\mathrm{A} \sigma\left(\frac{R s}{d}\right)^2 \mathrm{~T}^4\)=2AσT⁴

⇒ \(\frac{R s}{d}=\frac{1}{216}\)

∴ T′ = \(\frac{T}{(216)^{1 / 2} 2^{1 / 4}}\)

⇒ \(\frac{6000}{14.7 \times 1.189}\)

= 345K

∴ T′ = 70ºC

Question 15. Behaving like a black body sun emits maximum radiation at wavelength 0.48µm. The mean radius of the sun is 6.96 × 10⁸m. Stefan’s constant is 5.67 × 10^-8wm^-2k^-4and wien’s constant is 0.293 cm-k. The loss of mass per second by the emission of radiation from the sun is
Answer: Using Wien’s law

T = T = \(\frac{b}{\lambda_m}=\frac{0.293 \times 10^{-2}}{0.48 \times 10^{-6}}\)

= 6104 K

Energy given out by sun per second

= AσT⁴=4π (6.96 × 10⁸)²× 5.67 × 10^-8(6104)⁴

⇒ 49.285 × 10²⁵J

Loss of mass per second

m = \(\frac{E}{c^2}=\frac{49.285 \times 10^{25}}{9 \times 10^{16}}\)

m = 5.4×10⁹kg/s

Question 16. 50g of water and an equal volume of alcohol (relative density 0.8) are placed one after the other in the same calorimeter. They are found to cool from 60ºC to 50ºC in 2 minutes and 1 minute respectively if the water equivalent of the calorimeter is 2g then what is the specific heat of the alcohol?
Answer:

Given t_w= 2min, telco = 1 min

m_w= 50g, malco = 50 × 0.8 = 40g

S_w= 1 in cgs units, w = 2g

Therefore,

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{m_{\text {alco }}}\left\{\frac{t_{\text {alco }}}{t_w}\left[m_w+W\right]-W\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{1}{40}\left\{\frac{1}{2}[50+2]-2\right\}\)

⇒ \(\mathrm{S}_{\mathrm{alco}}=\frac{24}{50}\)

S_alco = 0.6 cgs units = 0.6 cal/gºC