Question

A 44.0 g sample of unknown metal at 99.0 °C was placed in a constant-pressure calorimeter, with a heat capacity of 12.4 J/°C, containing 80.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4 °C. Calculate the specific heat of the metal.

Answer 1

Here is what I have.

Invert 1
CO2(g) + 2H2O(l) → CH3OH(l) + 3/2O2(g) ΔHorxn = +726.4 kJ/mol
Multiply Equation 3 by 2
2H2(g) + O2(g) → 2H2O(l) ΔHorxn = 2 x (–285.8 kJ/mol) = -571.+6 kJ/mol
Add equations
CO2(g) + 2H2O(l) → CH3OH(l) + 3/2O2(g) ΔHorxn = +726.4 kJ/mol
2H2(g) + O2(g) → 2H2O(l) ΔHorxn = 2 x (–285.8 kJ/mol) = -571.+6 kJ/mol
CO2 (g) + 2H2O(l) + 2H2(g) + O2(g) --> CH3OH(l) + 3/2O2(g) + 2H2O(l)
ΔHorxn = +154.8 kJ/mol
Cancel common terms
CO2(g) + + 2H2(g) --> CH3OH(l) + 1/2O2(g) ΔHorxn = +154.8 kJ/mol
Add Equation 3
C(graphite) + O2(g) → CO2(g) ΔHorxn = –393.5 kJ/mol

Add both equations
CO2(g) + + 2H2(g) --> CH3OH(l) + 1/2O2(g) ΔHorxn = +154.8 kJ/mol
C(graphite) + O2(g) → CO2(g) ΔHorxn = –393.5 kJ/mol
CO2(g) + + 2H2(g) + C(graphite) + O2(g) --> CH3OH(l) + 1/2O2(g) +CO2(g) ΔHorxn = -238.7 kJ/mol
cancel terms
2H2(g) + C(graphite) + 1/2O2(g) --> CH3OH(l) ΔHorxn = -238.7 kJ/mol
Hf of C = 0
2H2(g) + 1/2O2(g) --> CH3OH(l) ΔHorxn = -238.7 kJ/mol

Cpmetal=-115.32

Answer 2

Ok let's see if I remember this... lol I would double check it but here it goes!
\[q_m=q_w\]
What I say here is the heat lost by the metal is equal to the heat gained by the water! so we substitute the equation for q, it's our calorimeter equation.
\[(Metal Side) mass*\Delta T*C_p=mass*\Delta T*C_p (Water Side)\]
\[(Metal Side) (44g)*(99-28.4)*(C_p)=(80g)*(28.4-24)*(12.4 J/C) (Water Side)\]
Solve for Cp of metal and find "1.405 J/C"

I would definately double check this cause we got totally different answers but I also haven't done this in forever lol.

Answer 3

Yeah IDK how to get there.

Answer 4

I got -1.405 J/C

Answer 5

nvm forgot to eliminate the negative sign.