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What mass of water at 10.0°C is needed to cool 500.0 mL of tea from 25.0°C to...

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kphilxo | Student, Undergraduate | eNotes Newbie

Posted September 25, 2012 at 8:44 PM via web

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What mass of water at 10.0°C is needed to cool 500.0 mL of tea from 25.0°C to 18.3°C? Assume that density and specific heat of the tea are the same as water.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 26, 2012 at 2:36 AM (Answer #1)

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The mass and density of tea is the same as water. Let the mass of water at 10.0 C needed to cool 500.0 mL of tea from 25.0 C to 18.3 C be X g. As the density of water is 1 g/ml, the mass of the tea to be cooled is 500 g.

The cool water can reduce the temperature of the tea till the two are at the same temperature. Once the two are at an equal temperature there is no change in temperature. This gives the equation:

`X*(18.3 - 10) = 500*(25 - 18.3)`

=> `8.3*X = 3350`

=> `X = 8350/8.3`

=> `X ~~ 403.61` g

The mass of water at 10 C required to cool the tea is 403.61 g.

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted September 26, 2012 at 12:47 AM (Answer #2)

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Heat must flow from Tea to Water, and as 18.3°C is the final temparature of Tea so it will also be the final temparature of water. Now let C be the specific heat of Tea & Water, and D be their density. Volume of water needed = Vw

Volume of Tea               = Vt = 500 mL

Mass of water needed = volume*density = Vw * D

Mass of Tea                                         = Vt * D

amount of Heat flow when temparature changes from T1 to T2 is

Q = mass * Specific heat * | T1 - T2 |

so heat lost by the Tea must be equal to heat gained by the water,

hence

(Vt * D)* C  * | 25 - 18.3 | = (Vw * D * C * |10 - 18.3 |

=> Vw = Vt * | 25 - 18.3 |/|10 - 18.3 |

          = 500 * 6.7/8.3

          =403.6 m L (approx)

so 403.61 mL water will be needed.          

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted September 26, 2012 at 12:49 AM (Answer #3)

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Heat must flow from Tea to Water, and as 18.3°C is the final temparature of Tea so it will also be the final temparature of water. Now let C be the specific heat of Tea & Water, and D be their density. Volume of water needed = Vw

Volume of Tea               = Vt = 500 mL

Mass of water needed = volume*density = Vw * D

Mass of Tea                                         = Vt * D

amount of Heat flow when temparature changes from T1 to T2 is

Q = mass * Specific heat * | T1 - T2 |

so heat lost by the Tea must be equal to heat gained by the water,

hence

(Vt * D)* C  * | 25 - 18.3 | = (Vw * D * C * |10 - 18.3 |

=> Vw = Vt * | 25 - 18.3 |/|10 - 18.3 |

          = 500 * 6.7/8.3

          =403.6 m L (approx)

so 403.61 mL water will be needed.    

 

so mass of water needed will be 

403.61 * 1 (density of water in CGS is 1 unit)

= 403.61 grams.

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