WBBSE Class 10 Physical Science Chapter 4 Thermal Phenomena Long Answer Questions

Class 10 Physical Science WBBSE Chapter 4 Thermal Phenomena Broad Answer Type Questions

Question 1. What is an anomalous expansion of water?
Answer:

Anomalous expansion of water: Usually liquids expand on heating. But in the case of water, we find a deviation from this general behaviour of the liquids within a certain range of temperature. The volume of water is minimum at 4°C and hence its density is maximum at 4°C. This phenomenon is called the anomalous expansion of water.

Question 2. What is thermo-metric conductivity?
Answer:

Thermometric conductivity: The rate of rise of temperature during the variable state is proportional to \(\frac{K}{P_s}\)(where K = thermal conductivity of the material, PS specific heat.)

This ratio is known as thermometric conductivity or the thermal conductivity per thermal capacity per unit volume.

Question 3. What is the fundamental principle of calorimetry? What are the conditions?
Answer:

Fundamental principle of calorimetry Heat lost by hot body = Heat gained by cold body.

Conditions:

• During the process of heat transfer, there is no heat exchange with the surroundings.
• No chemical reaction takes place between the bodies.

Question 4. Give a few applications of conductivity.
Answer:

Applications of conductivity:

Ice is packed in sawdust because air which is a bad conductor of heat being trapped in the sawdust prevents the transfer of heat from the surroundings to the ice. So, ice does not melt.

Question 5. Establish the relation: Coefficient of linear \((\Delta \mathrm{l})=\alpha \mathrm{l} \Delta \mathrm{t}\)
Answer:

Let us suppose that a rod has initial l₁ at t₁° and when it is heated to t₂, its length increases to l₂.

\(\begin{aligned}
& \left(l_2-l_1\right) \alpha l_1 \\
& \left(l_2-l_1\right) \alpha\left(t_2-t_1\right) \\
& \left(l_2-l_2\right) \alpha l_1\left(t_2-t_1\right) \\
& \left(l_2-l_1\right)=\alpha l_1\left(t_2-t_1\right)
\end{aligned}\)

(α = constant of proportionality)

This constant is called the coefficient of linear expansion of the solid.

\(\alpha=\frac{\mathrm{l}_2-\mathrm{b}_1 / \mathrm{l}_2}{\left(\mathrm{t}_2-\mathrm{t}_1\right)}\)

\(l_2=l_1\left\{1+\alpha\left(t_2-t_1\right)\right\}\)for a small increase in temperature At, if the length I of a rad increases by Al, then,

\(\begin{aligned}
& \alpha=\frac{\Delta \mathrm{l} / \mathrm{l}}{\Delta \mathrm{t}} \\
& \Delta \mathrm{l}=\alpha \mathrm{l} \Delta \mathrm{t}
\end{aligned}\)

WB Class 10 Physical Science Question Answer

Question 6. Establish the relation: ΔS = ΔβSΔt for superficial expansion.
Answer:

If S₁ be the initial surface area of a solid substance at t₁° and S₂, be that at t2 (t₂>t₁), the coefficient of superficial expansion is denoted by β.

\(\beta=\frac{\left(S_2-S_1\right) / S_1}{\left(t_2-t_1\right)}\) \(s_2=s_1\left\{1+\beta\left(t_2-t_1\right)\right\}\)

For a small increase in temperature Δt, if the surface area s of a solid increases by ΔS, then

\(\beta=\frac{\Delta s / s}{\Delta t}\)

Δs = βsΔt

Question 7. Establish the relation: ΔV = yVΔt of cubical expansion.
Answer:

If V₁ and V₂ are the volumes of a solid at t₁° and t₂°, (t₂>t₁) respectively, and y is the coefficient of cubical expansion.

\(y=\frac{\left(v_2-v_1\right) / v_1}{\left(t_2-t_1\right)}\)

V₂ = V₁ {1+y (t₂-t₁)}

For a small rise in temperature Δt, if the volume V of a solid increases by ΔV

\(y=\frac{\Delta V / V}{\Delta t}\)

ΔV=YVΔt

Question 8. Establish the relation : \(V_t=V_0\left(1+\frac{t}{273}\right)\) for Charle’s law.
Answer:

Let, Vo denote the volume of a given mass of a gas at 0°C.

V_t = The volume of the same mass of gas at t°C.

The pressure in both cases is the same.

For Charle’s law, the increase in volume of the gas

= \(\frac{\text { volume at } 0^{\circ} \mathrm{C}}{273} \times \text { rise in temp }=v_t-v_0=\frac{v_0}{273} \times t\)

\(v_t=v_0(1+y p t)\)

Question 9. Establish the relation V_t = V₀ (1+ y_pt) for the volume coefficient of a gas.
Answer:

If V₀ and V₂ are the volume of a given mass of a gas at 0°C and t°C respectively.

\(\mathrm{y}_{\mathrm{p}}=\frac{\text { fractional increase in volume }}{\text { rise of temperature }}=\frac{\left(\mathrm{v}_{\mathrm{t}}-\mathrm{v}_{\mathrm{o}}\right) / \mathrm{v}_{\mathrm{o}}}{\mathrm{t}}\)

V_t = V₀ (1+ y_pt)

Question 10. The thickness of an iron plate is 4mm. and its area is 150 sqm. The temperature of the two sides is 100°C, 30°C and 3940. Caloric heat is conducted in one second from one side to the other. Determine the thermal conductivity of heat.
Answer:

Given

The thickness of an iron plate is 4mm. and its area is 150 sqm. The temperature of the two sides is 100°C, 30°C and 3940. Caloric heat is conducted in one second from one side to the other.

\(\frac{\mathrm{Q}}{\mathrm{t}}=3940 \mathrm{cal} / \mathrm{s}\) , A = 150 cm^{2 ,}

x = 4mm = 0.4 cm², θ₁ = 30°C, θ₂ = 100°C

Thermal conductivity of iron.

\(\mathrm{K}=\frac{\mathrm{Qx} / \mathrm{t}}{\mathrm{A}\left(\theta_2-\theta_1\right)}=\frac{3940 \times 0.4}{150(100-30)} \text { c.g.s unit. }=0.15 \text { c.g.s. unit. }\)

Question 11. The thermal conductivity of aluminium is 0.50 cal/sec. cm degree. Express the same in units of watt/m degree.
Answer:

Given

The thermal conductivity of aluminium is 0.50 cal/sec. cm degree.

K 0.5 cal/5 cm degree

= 0.5 x 4.2 Joule/s Cm degree

= 0.5 × 4.2 × 100 Joule/s m degree

= 210 watt/m degree.

Question 12. A pond is covered with ice 5cm. thick and the temperature of overlying air is -10°C. Find the rate of conduction of heat by the ice per unit square centimetre area. (Given the Thermal conductivity of ice) = 5 x 10^-3 c.g.s unit)
Answer:

Given

A pond is covered with ice 5cm. thick and the temperature of overlying air is -10°C.

k =5 x 10^-3 c.g.s unit, A = 1 cm²,

x = 5cm, θ₂-θ₁, 0-(-10)

or 10°C.

Rate conduction of heat \(\frac{Q}{t}\)

\(=\frac{\mathrm{KA}\left(\theta_2-\theta_1\right)}{\mathrm{X}}=\frac{5 \times 10^{-3} \times 1 \times 10}{5} \mathrm{cal} / \mathrm{s} .\)

Physics Class 10 WBBSE Question 13. At what temperature will the volume of a given mass of gas be double of what is at 30°C, if the pressure remains constant?
Answer:

Let, t°C = the required temperature

Applying Charles’s law,

\(\frac{\mathrm{V}}{\mathrm{V}^1}=2=\frac{273+\mathrm{t}}{273+30}\) \(2=\frac{273+t}{303}\)

273+ t = 606

t=333°C

Question 14. The volume of a liquid is 830 m³ at 30°C and it is 850 m3 at 90°C. Find the coefficient of volume expansion of the liquid.
Answer:

Given

The volume of a liquid is 830 m³ at 30°C and it is 850 m3 at 90°C.

If V₁ and V₂ are the volume of a liquid at temperatures t1°C and t°C respectively.

\(y=\frac{\text { Increase in volume }}{\text { Original volume } \times \text { Rise in temperature }}=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)

[V₁ = 830m³, t₁= 30C, V₂ = 850m3, t² = 90c]

\(y=\frac{850-830}{830(90-30)}=\frac{20}{830 \times 60}=4 \times 10^{-4 /{ }^{\circ} \mathrm{C}}\)

Physics Class 10 WBBSE Question 15. The coefficient of cubical expansion of as metal is 7.2×10-5 °C^-1. Find the

1. Coefficient of linear expansion
2. Coefficient of superficial expansion. (Given = 7.2 × 10-5/°C)
   Answer:

1. Coefficient linear expansion

\(\alpha=\frac{1}{3} y=\frac{1}{3} \times 7.2 \times 10^{-5}=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

2. Coefficient of superficial expansion

\(\beta=\frac{2}{3} y=\frac{2}{3} \times 7.2 \times 10^{-5}=4.8 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

Question 16. The base of an iron saucepan has a diameter of 20 cm at 15°C. What will be the increased area of the base of the saucepan when it is filled with boiling water? (Given the coefficient of linear expansion of iron 12 x 10^-6 °C)
Answer:

Given

The base of an iron saucepan has a diameter of 20 cm at 15°C.

The radius of the base = 20/2 cm = 10 cm.

S₂ – S₁ = S₁ x B x (t₂ – t₁), the formula is necessary to define the coefficient of surface expansion.

S₁ = π x 10² Cm², β = 2α = 2 x 12 × 10^-6 per°C

(t₂ – t₁) (100 – 15)°C = 85°C

the required increase in area = (π x 10² x 24 x 10-6 × 85)

= 0.6404 Cm².

Question 17. The coefficient of apparent expansion of a liquid is 18 x 10-5°C when an iron vessel is used and 14.46 x 10-5°C when an aluminium vessel is used. If the coefficient of iron be 12 x 10-6/0C. Find that of aluminium.
Answer:

Given

The coefficient of apparent expansion of a liquid is 18 x 10-5°C when an iron vessel is used and 14.46 x 10-5°C when an aluminium vessel is used. If the coefficient of iron be 12 x 10-6/0C.

The coefficient of volume expansion of iron = 8 × 12 × 10^-6 °C

= 3.6 × 10^-5 °C.

The coefficient of real expansion of the liquid is given by

Yr = (18 x 10-5 +3.6 x10^-5 )/°C

= 21.6 x 10^-5C

If the coefficient of linear expansion of aluminium is a, then

21.6 x 10^-5 = 14.46 x 10^-5 + 3α

\(\alpha=\frac{7.14}{3} \times 10^{-5 /{ }^{\circ} \mathrm{C}}\)

= 2.33 x 10^-5/°C