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OWLv2 x e cvg.cengagenowcom/ilrn/takeAssignment/takeC a tr o 2 I Save and CHAPTER 5-PRINCIPLES OF CHEMICAL REACTIVITY: ENERGYAND CHEMICAL REACTIONS Study OPrevious Page 4 of 12 Next O Submit Qui Use the References to access important values if needed for this question. A 55.5 -g piece of tin at 95.0 "Cwas immersed into 65.0 g water at 25.0 C. The water temperature rose to 28.1 "C. Calculate the molar heat capacity of the metal. 24.7 J/mol C o 113 J/mol. C o 0.227 J/mol "C 0 6.44 J/mol "C o 27.0 J/mol "C

Explanation / Answer

Let substance 1 be water and substance 2 be tin

m1 = 65.0 g

T1 = 25.0 oC

C1 = 4.184 J/goC

m2 = 55.5 g

T2 = 95.0 oC

C2 = to be calculated

Final temperature,

T = 28.1 J/goC

use:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

55.5*C2*(95.0-28.1) = 65.0*4.184*(28.1-25.0)

C2= 0.227 J/goC

Answer: 0.227 J/goC

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