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physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

11.18 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 vaporization of water = 2256 × 10³J kg^–1.

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Payal Gupta | Contributor-Level 10

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11.18 Base area of the boiler, A = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Let us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 s

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

The amount of heat flowing into water through the brass base of the boiler is given by:

$θ$ = $\frac{K A (T_{1} - T_{2}) t}{l}$ , where

$T_{1}$ = Flame temperature in contact with the boiler

$T_{2}$ = Boiling point of water = 100 $?$

Heat required for boiling water $θ$ = mL, where L = heat of vaporization of water = 2256 × 103 J kg–1

By equating for $θ$ we g

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11.18 Base area of the boiler, A = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Let us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 s

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

The amount of heat flowing into water through the brass base of the boiler is given by:

$θ$ = $\frac{K A (T_{1} - T_{2}) t}{l}$ , where

$T_{1}$ = Flame temperature in contact with the boiler

$T_{2}$ = Boiling point of water = 100 $?$

Heat required for boiling water $θ$ = mL, where L = heat of vaporization of water = 2256 × 103 J kg–1

By equating for $θ$ we get

6 $\times 2256 \times 10^{3}$ = 109 $\times 0.15 \times (T_{1} - T_{2}) \times 60 / 0.01$

$(T_{1} - T_{2})$ = 137.98 $?$

$T_{1}$ = 237.98 $? (F l a m e t e m p e r a t u r e i n c o n t a c t w i t h t h e b o i l e r)$

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11.18 Base area of the boiler, A = 0.15 m2Thickness of the boiler, l = 1.0 cm = 0.01 mBoiling rate of water, R = 6.0 kg/minLet us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 sThermal conductivity of brass, K = 109 J s–1 m–1 K–1The amount of heat flowing into water through the brass base of the boiler is given by:<math><mi>θ</mi></math> = <math><mfrac><mrow><mrow><mi>K</mi><mi>A</mi><mfenced separators="|"><mrow><mrow><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow></mfenced><mi>t</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></mfrac></math> , where<math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = Flame temperature in contact with the boiler<math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = Boiling point of water = 100 <math><mi>?</mi></math>Heat required for boiling water <math><mi>θ</mi></math> = mL, where L = heat of vaporization of water = 2256 × 103 J kg–1By equating for <math><mi>θ</mi></math> we get6 <math><mo>×</mo><mn>2256</mn><mi></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> = 109 <math><mo>×</mo><mn>0.15</mn><mo>×</mo><mfenced separators="|"><mrow><mrow><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow></mfenced><mi></mi><mo>×</mo><mn>60</mn><mo>/</mo><mn>0.01</mn></math><math><mfenced separators="|"><mrow><mrow><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow></mfenced></math> = 137.98 <math><mi>?</mi></math><math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 237.98 <math><mi>?</mi><mi></mi><mo>(</mo><mi mathvariant="normal">F</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">w</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">b</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi><mo>)</mo></math>

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At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter. 6.241cm? Both the diameters have been measured at room temperature (27°C).

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alok kumar singh

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A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

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(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

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This is a multiple choice answer as classified in NCERT Exemplar

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This is a multiple choice answer as classified in NCERT Exemplar

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