Answer:
c = 1,100 J/kgºC
Explanation:
Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the unknown metal.
The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:
Q = c * m* Δt
where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.
In our case, we can write the following equality:
(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (cₓ*mₓ*Δtₓ)
Replacing by the givens , and taking cAl = 0.9 J/gºC, we have:
Qg= 0.9 J/gºC*97g*10ºC + 4.186 J/gºC*259g*10ºC = 11,715 J(1)
Ql = cₓ*159g*67ºC (2)
Based on all the previous assumptions, we have:
Qg = Ql
So, we can solve for cx, as follows:
cx = 11,715 J / 159g*67ºC = 1.1 J/gºC (3)
Expressing (3) in J/kgºC:
1.1 J/gºC * (1,000g/1 kg) = 1,100 J/kgºC
