A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 75°C. The other sample has a mass of 66 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample. (Pick the answer closest to the true value.)A. 1950 joules Co/kgB. 975 joules Co/kgC. 3950 joules Co/kgD. 250 joules Co/kgE. 8500 joules Co/kg

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 copper and 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) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC