A Calorimetry ProblemI have a question in calorimetry from an old competitive exam. The question is:The temperature of 100 grams of water is to be raised from 24 ∘ C to 90 ∘ C by adding steam to it. Find the mass of steam required.I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the 'watered steam' will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!I would appreciate any help on this matter. Thank you :)

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A Calorimetry ProblemI have a question in calorimetry from an old competitive exam. The question is:The temperature of   100 grams of water is to be raised from       24    ∘  C to       90    ∘  C by adding steam to it. Find the mass of steam required.I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the &#039;watered steam&#039; will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!I would appreciate any help on this matter. Thank you :)

Leroy Lowery · Accepted Answer

Your method seems correct. Here are the details:      Q          H      e      a      t      W      a      t      e      r        =  m  ×  c  ×  (      T    2    −      T    1    )  =  100  ×  1  ×  66  =  6600 caloriesFrom 1 gram of steam,  Q  =      L    v    +  c  ×  (  100  −      T    2    )  =  540  +  1  ×  10  =  550Therefore, grams of steam needed   =      6600    550    =  12 grams

A Calorimetry Problem I have a question in calorimetry from an old competitive exam. The question

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