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A student brought 25.50 g of iron ( Cp = 0.449 J/(g°C)) at 23.8°C into contact w

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A student brought 25.50 g of iron ( Cp = 0.449 J/(g°C)) at 23.8°C into contact with 54.44 g of silver ( Cp = 0.235 J/(g°C)) at 70.5°C. Determine the equilibrium temperature ( to 0.1 °C)

Mar 18th, 2015

Q=mc DT

for iron, q= 25.5*0.449x=11.45x

for silver, q=54.44*0.235x1=12.79x1

q1=q2

12.79/11.45=1.11

(Tf-70.5)/(Tf-23.8)=1.11

Tf-70.5=1.11Tf-26.42

Tf= 44.08= 44.1 degree celcius.......................


Mar 18th, 2015

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