100ml of water initially at 35°c is placed inside a calorimeter. what will be the final temperature in the calorimeter if 50g of aluminum initially at 100°c is added to the water?

51.27°C

We are given;

Density of water is 1 g/mL, therefore, Mass is 100 g

We are required to calculate the final temperature of the mixture;

We are going to use the following steps;

Quantity of heat absorbed = mcΔT

Q = mcΔT

Assuming the final temperature is X

Then, ΔT is (X-35)°C

But, specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 100 g × 4.184 J/g°C ×  (X-35)°C

  = 418.4X - 14,644 Joules

Mass of aluminium is 50 g

Specific heat capacity of aluminium is 0.90 J/g°C

Assuming the final temperature is X, thus ΔT is (100 - X)°C

Thus;

Q = 50 g × 0.9 J/g°C × (100 - X)

   = 4500 - 45X Joules

We need to know that the Quantity of heat absorbed is equal to the quantity of heat released.

Therefore;

418.4X - 14,644 J =  4500 - 45X J

373.4X = 19,144

         X = 51.27°C

Therefore, the final temperature of the mixture is 51.27°C

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.