11.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]

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Payal Gupta | Contributor-Level 10

5 months ago

11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m

Thickness of the icebox, l = 5 cm = 0.05 m

Mass of ice kept in the box, m = 4 kg

Time gap, t = 6 h = 6 $\times 60 \times 60$ s

Outside temperature, T = 45 °C

Coefficient of thermal conductivity of thermocole, K = 0.01 J $s^{- 1} m^{- 1} K^{- 1}$

Heat of fusion of water, L = 335 $\times 10^{3}$ J ${k g}^{- 1}$

Let m’ be the mass of the ice melts in 6 h

The amount of heat lost by the food: $θ$ = $\frac{K A (T - 0) t}{l}$ , where

A = Surface area of the box = 6 $s^{2}$ = 6 $\times {0.3}^{2} m^{2}$ = 0.54 $m^{2}$

$θ$ = $\frac{0.01 \times 0.54 (45 - 0) \times 6 \times 60 \times 60}{0.05}$ = 104976 J

We also know $θ = m' L$ so m’ = 104976/ (335 $\times 10^{3})$ = 0.313 kg

Hence the amount of ice remains

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11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m

Thickness of the icebox, l = 5 cm = 0.05 m

Mass of ice kept in the box, m = 4 kg

Time gap, t = 6 h = 6 $\times 60 \times 60$ s

Outside temperature, T = 45 °C

Coefficient of thermal conductivity of thermocole, K = 0.01 J $s^{- 1} m^{- 1} K^{- 1}$

Heat of fusion of water, L = 335 $\times 10^{3}$ J ${k g}^{- 1}$

Let m’ be the mass of the ice melts in 6 h

The amount of heat lost by the food: $θ$ = $\frac{K A (T - 0) t}{l}$ , where

A = Surface area of the box = 6 $s^{2}$ = 6 $\times {0.3}^{2} m^{2}$ = 0.54 $m^{2}$

$θ$ = $\frac{0.01 \times 0.54 (45 - 0) \times 6 \times 60 \times 60}{0.05}$ = 104976 J

We also know $θ = m' L$ so m’ = 104976/ (335 $\times 10^{3})$ = 0.313 kg

Hence the amount of ice remains after 6 h = 4 – 0.313 = 3.687 kg

less

11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 mThickness of the icebox, l = 5 cm = 0.05 mMass of ice kept in the box, m = 4 kgTime gap, t = 6 h = 6 <math><mo>×</mo><mn>60</mn><mo>×</mo><mn>60</mn></math> sOutside temperature, T = 45 °CCoefficient of thermal conductivity of thermocole, K = 0.01 J <math><msup><mrow><mrow><mi>s</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup><msup><mrow><mrow><mi>K</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math>Heat of fusion of water, L = 335 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> J <math><msup><mrow><mrow><mi>k</mi><mi>g</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math>Let m’ be the mass of the ice melts in 6 hThe amount of heat lost by the food: <math><mi>θ</mi></math> = <math><mfrac><mrow><mrow><mi>K</mi><mi>A</mi><mfenced separators="|"><mrow><mrow><mi>T</mi><mo>-</mo><mn>0</mn></mrow></mrow></mfenced><mi>t</mi></mrow></mrow><mrow><mrow><mi>l</mi></mrow></mrow></mfrac></math> , whereA = Surface area of the box = 6 <math><msup><mrow><mrow><mi>s</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 6 <math><mo>×</mo><msup><mrow><mrow><mn>0.3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 0.54 <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math><math><mi>θ</mi></math> = <math><mfrac><mrow><mrow><mn>0.01</mn><mo>×</mo><mn>0.54</mn><mfenced separators="|"><mrow><mrow><mn>45</mn><mo>-</mo><mn>0</mn></mrow></mrow></mfenced><mo>×</mo><mn>6</mn><mi mathvariant="normal"></mi><mo>×</mo><mn>60</mn><mo>×</mo><mn>60</mn></mrow></mrow><mrow><mrow><mn>0.05</mn></mrow></mrow></mfrac></math> = 104976 JWe also know <math><mi>θ</mi><mo>=</mo><mi>m</mi><mi>'</mi><mi>L</mi></math> so m’ = 104976/ (335 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup><mo>)</mo></math> = 0.313 kgHence the amount of ice remains after 6 h = 4 – 0.313 = 3.687 kg

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