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

nuhygjhuhuouh9oygyuhuh8h9guig97uogh8uog978o0rfyhkmkv h