A student brought 26.70 g of iron ( Cp = 0.449 J/(g°C)) at 25.0°C into contact w

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A student brought 26.70 g of iron ( Cp = 0.449 J/(g°C)) at 25.0°C into contact with 51.56 g of silver ( Cp = 0.235 J/(g°C)) at 67.0°C. Determine the equilibrium temperature ( to 0.1 °C)

Feb 1st, 2015

With no change in state:

Q1=C1m1(Tf-T1i)   Q2=C2m2(Tf-T2i) Equilibrium indicates that the heat flowing out of the higher temperature object must equal the heat flowing into the lower temperature object. since negative indicate heat flowing out and positive indicate heat adding in, (and since the value is the same) the two should add up to cancel each other out, giving 0. Tf is the answer we want, final temperature.

So we set the 2 equations up: With 1 denoting iron and 2 denoting silver

C2m2r(Tf-T2i) + C1m1(Tf-T1i)=0

Plugging numbers in we get: (0.235)(51.56)(Tf-67) + (0.449) (26.7)(Tf-25)=0

then we arrange to solve for Tf

12.12(Tf-67) + 11.99(Tf-25)=0

12.12Tf- 812.04+ 11.99Tf- 299.75=0

24.11Tf - 1111.79=0

Tf=46.1 Celsius

:)

Feb 2nd, 2015

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