NCERT solution Class 11 thermal properties of matter

NCERT Solutions

Class 11 Physics

Thermal Properties of Matter

NCERT Questions

Q. 10.1
Q. 10.2
Q. 10.3
Q. 10.4
Q. 10.5
Q. 10.6
Q. 10.7
Q. 10.8
Q. 10.9
Q. 10.10
Q. 10.11
Q. 10.12
Q. 10.13
Q. 10.14
Q. 10.15
Q. 10.16
Q. 10.17
Q. 10.18
Q. 10.19
Q. 10.20

Q. 10.18

A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^–1 m^–1 K^–1; Heat of vaporisation of water = 2256 × 10³ J kg^–1.

Given:

Thermal conductivity of brass K = 109 J s^–1 m^–1 K^–1

Area of brass boiler, A = 0.15m²

Thickness d = 1.0cm = 10^-2m

Boiling rate of water, R =0.6 kg/min

Time, t = 1 min = 60sec

Heat of vaporisation of water = 2256 × 10³ J kg^–1

The amount of heat flowing into water through the brass bas of the boiler is given as:

θ = (KA (T₁ – T₂)/t)/d … (i)

Temperature of the flame in contact with the boiler= T₁

Boiling point of water T₂= 100°C

Therefore, heat required for boiling water, θ = mL ... (ii)

Equating equations (i) and (ii)

=> mL =(KA (T₁ – T₂)/t)/d

(T₁ – T₂) = (mLd)/ (KAt)

Therefore, (T₁ – 100) = (6 x 2556 x 10³ x 0.01)/ (109 x 0.15 x 60)

(T₁ – 100) = 137.98

= 138⁰C

0r = 138 + 100

= 238 ⁰C

Therefore, the temperature of the part of the flame in contact with the boiler is

238 ⁰C.

Video solution:

Thermal Properties of Matter Class 11 Physics NCERT Chapter 10 NCERT Solutions

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