# A 36.07 g sample of a substance is initially at 27.8°C. After absorbing 2639 J of heat, the temperature of the substance is 109.0°C. What is the specific heat of the substance?

Jul 25, 2016

I found: $c = 0.9 \frac{J}{{g}^{\circ} C}$

#### Explanation:

I would use the relationship between heat $Q$, mass $m$, specific heat $c$ and change in temperature $\Delta T$ as:
$Q = m c \Delta T$
so:
$2639 = 36.07 \cdot c \cdot \left(109.0 - 27.8\right)$
rearranging:
$c = 0.9 \frac{J}{{g}^{\circ} C}$

Aluminum perhaps?