A 54.7-g sample of aluminum at 95.0°C is dropped into 35.0 g of water at 40.0°C. What is the final temperature of the mixture? (specific heat capacity of aluminum = 0.89 J/g°C; specific heat capacity of water = 4.184 J/g°C)

Answer 1

Summation of all heat is zero.
Q,aluminum + Q,water = 0

The formula for Q is
Q = mc(T2-T1)
where
m = mass (g)
c = specific heat capacity (J/g-K)
T = temperature (K)

Substitute the values. Note that the final Temperature (T2) is the same for both:
[mc(T2-T1)],aluminum + [mc(T2-T1)],water = 0
54.7 * 0.89 * (T2 - 95) + 35 * 4.184 * (T2 - 40) = 0

Now solve for T2. The value you should get must be between 40 and 95 °C.
Hope this helps~ `u`

Answer 2

23°C