trimblewg trimblewg
  • 25-11-2020
  • Chemistry
contestada

Please help on the image below.

Please help on the image below class=

Respuesta :

ardni313
ardni313 ardni313
  • 27-11-2020

Heat capacity of Calorimeter = 10.7 J/°c

Further explanation

Heat lost=Heat gained

Q in = Q out

-Q lost(hot water)=Q gained (cold water+calorimeter)

-m.c.Δt=m.c.Δt+C.Δt

[tex]\tt \rightarrow -50\times 4.18\times (32.7-43)=50\times 4.18\times (32.7-22.9)+(32.7-22.9)\times C_{cal}\\\\\rightarrow 2152.7=2048.2+9.8\times C_{cal}\\\\104.5~J=9.8\times C_{cal}\\\\C_{cal}=\dfrac{104.5}{9.8}=10.7~J/^oC[/tex]

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