# An object with a mass of 24 g is dropped into 6400 mL of water at 0^@C. If the object cools by 60 ^@C and the water warms by 16 ^@C, what is the specific heat of the material that the object is made of?

$m \cdot c \cdot \Delta \left(t\right)$ equation can be used. For water $6.4 \cdot 16 \cdot 1$ will be equal to $0.024 \cdot 60 \cdot C$. When you solve this, you will get $C = 71.11 \frac{k c a l}{k g \cdot \mathrm{de} g r e e s C}$