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

Explanation:

We are given;

Volume of water is 100 mL

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

Initial temperature is 35°C Mass of the calorimeter is 50 g Initial temperature of aluminium is 100°C

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

We are going to use the following steps;

Step 1: Calculate the amount of heat absorbed by water

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

Step 2: Calculate the quantity of heat released by Aluminium

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

Step 3: Calculate the final temperature, X°C

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