. A 60.0 g piece of Zn at 20.0 o C is dropped into 200.0 g of water at 100.0 o C. Determine the final temperature reached by the Zn and water, knowing that the specific heat capacity of Zn and water are 0.389 J/(g. o C) and 4.184 J/(g. o C), respectively.

97.8 C

Explanation:

Recall the equation for specific heat:

q = mcΔT

There are several things to know in order to answer this question:

1. Whatever heat energy one object gains is the same as the heat energy the other has lost (conservation of energy).

2. Since one substance loses heat and the other gains heat, you need to multiply one side by -1 to account for the opposite direction of heat transfer.

3. The final temperature of both substance will be the same (zeroth law of thermodynamics)

Below, the left side of the equation represents water and the right side represents zinc.

-1 x 200 x 4.184 x (T - 100) = 60 x 0.389 x (T - 20)

Solve this equation and you get T = 97.8.