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o-na [289]
3 years ago
7

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 initiall

y 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.
Chemistry
2 answers:
irakobra [83]3 years ago
4 0
<h3><u>Answer;</u></h3>

<em>-49 °C</em>

<h3><u>Explanation and solution;</u></h3>
  • 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

<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol

                                 =1.17 kJ => 1170 J  

<em>But heat lost =heat gained</em>

<em>Therefore;</em> 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 <em>at two significant figures)</em>

<em>Hence</em>, final temperature = 49 °C

Amanda [17]3 years ago
4 0

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

(2257 J/g) x (0.48 g) = 1083.36 J </span><span>

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

25°C + 21.81°C = 47°C</span><span>

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

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<h3>Further explanation</h3>

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I think so... I'm currently learning this too but you should be correct
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