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A drop of water weighing 0.48 g is vaporized at 100 ?c and condenses on the surface of a 55-g block of aluminum that is initially at 25 ?c. if the heat released during condensation goes only toward heating the metal, what is the final temperature in celsius of the metal block? (the specific heat capacity of aluminum is 0.903 j/g ?c.

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BlueSky06

Hi, the solution would be like this for this specific problem:

(2257 J/g) x (0.48 g) = 1083.36 J 

(1083.36 J) / (0.903 J/g ∘C) / (55 g) = 21.81°C change 

25°C + 21.81°C = 47°C

I hope this helps and if you have any further questions, please don’t hesitate to ask again. 

Answer;

-49 °C

Explanation and solution;

  • Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.  

Moles water = 0.48 g / 18.02 g/mol

                      =0.0266  moles

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

                                 =1.17 kJ => 1170 J  

But heat lost =heat gained

Therefore; Heat gained by aluminium = 1170 J  

        1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242  

                   1170 + 1242 = 49.7 T  

            T = 48.5 °C ( 49 °C at two significant figures)

Hence, final temperature = 49 °C

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