5. An aluminum cup of mass 46 g contains 150 g of water. The cup and water
are initially in thermodynamic equilibrium at room temperature (23 C). A 50-
g sample of an unknown alloy is heated to a temperature of 350 °C and
dropped into the water. The entire system (cup, water, alloy) is sealed in an
insulated container and allowed to come to equilibrium, at which point its
temperature is found to be 45°C. What is the specific heat of the alloy?
Hint: J
alloy - J + J Al cap

C(alloy) = 0.965 J/g°C

Explanation:

In the problem, the Heat of the alloy = - (Heat of water + Heat of Calorimeter). This, could be written as:

m×C(alloy)×ΔT = - ( m×C(H₂O)×ΔT +m×C(Al)×ΔT)

Replacing knowing C(H₂O) = 4.184 J/g°C and C(Al) = 0.900 J/g°C:

50g×C(alloy)×(45°C - 350°C) = - ( 150g×4.184J/g°C×(45°C - 23°C) +46g×0.900J/g°C×(45°C-23°C))

15250g°C×C(alloy) = 14718J

C(alloy) = 0.965 J/g°C

I hope it helps!

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.