Enthalpy question?

Question

a 10.4 g sample of unknown metal at 99.0 C was placed in a constant pressure calorimeter containing 75.0 g of water at 23.5 C. The final temp of the system was 25.7. Calculate the specific heat of the metal.

Answer 1

Q= mc(delta T)

Answer 2

ok.  So the metal is at a temperature of 99.0 C and the water is at a temperature of 23.5 C.   It should be clear that the heat lost by the metal should be exactly the same but opposite in sign of the heat gained by the water.

Written out, that is:  

qwater + qmetal = 0   or you can write it qwater = -qmetal where q = m*c*(Delta T).  

qwater = (75.0g)cwater(25.7 - 23.5), the c, or specific heat,  for water depends on what units we are looking for.

cwater = 1 calorie/gram °C = 4.186 joule/gram °C

generally enthalpy is expressed in joules so I will use the 4.186 one.

qwater = (75.0)(4.186)(25.7-23.5)

qwater = 690.69 J

so we know qwater + qmetal = 0

690.69 + (cmetal)(mass of metal)(Delta T for metal) = 0

690.69 + (cmetal)(10.4)(-73.3) = 0     (delta T for metal is negative because it lost heat to the surroundings).

cmetal = -690.69/(10.4 * -73.3)

cmetal = 0.906 joule/gram °C

And now we can see qualitatively if this makes sense:

we began with substantially less unknown metal and the metal changed temperature much more then the water indicating that it should have a much lower specific heat then water which it does.

Answer 3

54.3?