Question

In the experiment to measure specific heat of a metal, you get the following data. What is the specific heat of the metal in J/kg.C?
(answer within +-3% will considered correct) MASSES: water= 259 g, metal=159 g, calorimeter=97 g INITIAL TEMPERATURES: water+calorimeter=21ºC, metal=98ºC Final equilibrium temp=31ºC Calorimeter is made of aluminum.

Answer 1

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