A26.6 g sample of aluminum at 100.4 °c is added to 100.6 g of water at 21.5 °c in a constant pressure calorimeter. what is the final temperature of the water in °c? the specific heat capacity of aluminum is 0.903 j/goc.

Given:
m₁ = 26.6 g, the mass of aluminum
T₁ = 100.4 °C, the temperature of the aluminum
c₁ = 0.903 J/(g-°C), the specific heat of aluminum

m₂ = 100.6 g, the mass of water
T₂ = 21.5 °C, the temperature of water
The specific heat of water is c₂ = 4.184 J/(g-C)

Let T =  the final temperature of the water and aluminum.
Calculate heat loss of the aluminum.
Q₁ = (26.6 g)*(0.903 J/(g-C))*(100.4-T C) = 24.1098(100.4 - T)  J

Calculate heat gained by the water
Q₂ = (100.6 g)*(4.184 J/(g-C))*(T - 21.5 C) = 420.1904(T - 21.5) J

Conservation of energy requires that Q₁ = Q₂.
Therefore
420.1904(T - 21.5) = 24.1098(100.4 - T)
444.3T = 1.1455 x 10⁴
T = 25.782 °C

25.8 °C

Me i need

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.