$\overline{)\mathbf{q}\mathbf{=}\mathbf{mc}\mathbf{\u2206}\mathbf{T}}$

c_{Al} = 0.900 J/g°C

**Convert q in J:**

$\mathbf{q}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{}\overline{)\mathbf{kJ}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kJ}}}$ **= 2400 J**

**Calculate the final temperature (T _{f}):**

Suppose that 26 g of each of the following substances is initially at 28.0 ºC. What is the final temperature of each substance upon absorbing 2.40 kJ of heat?

a) aluminum

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