Question

A 2.64E+1 g sample of aluminum at 1.006E+2 °C is added to 1.004E+2 g of water at 2.20E+1 °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/g°C.

A. (40.5 °C)
B. (23.7 °C)
C. (31.2 °C)
D. (52.8 °C)

Answer 1

To find the final temperature of the water, calculate the heat gained by the water and the heat lost by the aluminum using the principle of conservation of energy. Set the two equations equal to each other and solve for ΔTwater.

Step-by-step explanation:

We can use the principle of conservation of energy to solve this problem. The heat gained by the water is equal to the heat lost by the aluminum.

First, calculate the heat gained by the water:

Qwater = mwater * cwater * ΔTwater

where mwater = 1.004E+2 g, cwater = 4.184 J/g°C (specific heat of water), and ΔTwater is the change in temperature for the water.

Next, calculate the heat lost by the aluminum:

Qaluminum = maluminum * caluminum * ΔTaluminum

where maluminum = 2.64E+1 g, caluminum = 0.903 J/g°C (specific heat of aluminum), and ΔTaluminum is the change in temperature for the aluminum.

Since no heat is transferred to the surroundings, the heat gained by the water is equal to the heat lost by the aluminum. Therefore,

Qwater = Qaluminum

Substituting the values, we can solve for the final temperature of the water, ΔTwater.