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

5 months ago

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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