**AP Physics 2 Exam Question:**

1. A 200 g aluminum block initially at 80°C is placed in 400 g of water initially at 20°C. The final temperature of the mixture is 30°C. Assuming no heat is lost to the surroundings, calculate the specific heat capacity of the aluminum block.

a) 0.9 J/g°C
b) 0.4 J/g°C
c) 0.2 J/g°C
d) 0.5 J/g°C
e) 0.1 J/g°C

**Answer:**

To solve this problem, we can use the equation for heat transfer:

\[ Q = mcΔT \]

Where Q is the heat lost by the aluminum block, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

We can set the heat lost by the aluminum block equal to the heat gained by the water:

\[ mcΔT = mcΔT \]

\[ (200g)(c_{Al})(80 - 30) = (400g)(4.18 J/g°C)(30 - 20) \]

\[ (200g)(c_{Al})(50) = (400g)(4.18 J/g°C)(10) \]

Solving for \( c_{Al} \):

\[ c_{Al} = \frac{(400g)(4.18 J/g°C)(10)}{(200g)(50)} \]

\[ c_{Al} = 0.418 J/g°C \]

Thus, the specific heat capacity of the aluminum block is approximately 0.4 J/g°C (b).

Therefore, the correct answer is b) 0.4 J/g°C.