# How do you calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.75 * 10^4 joules of heat, and its temperature changes from 32°C to 57°C?

Jan 29, 2016

1.8"J"/("g" ""^@"C")

#### Explanation:

A substance's specific heat tells you how much heat much either be added or removed from $\text{1 g}$ of that substance in order to cause a ${1}^{\circ} \text{C}$ change in temperature.

The equation that establishes a relationship between specific heat, heat added or removed, and change in temperature looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat added / removed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature

In your case, the $\text{1500.0-g}$ piece of wood is said to absorb a total of $6.75 \cdot {10}^{4} \text{J}$ of heat. This caused its temperature to increase from ${32}^{\circ} \text{C}$ to ${57}^{\circ} \text{C}$.

The difference between the final temperature and the initial temperature of the sample will be the value for $\Delta T$.

$\Delta T = {57}^{\circ} \text{C" - 32^@"C" = 25^@"C}$

This means that the specific heat of the wood is equal to

$q = m \cdot c \cdot \Delta T \implies c = \frac{q}{m \cdot \Delta T}$

Plug in your values to get

c = (6.75 * 10^4"J")/("1500.0 g" * 25^@"C") = color(green)(1.8"J"/("g" ""^@"C"))

The answer is rounded to two sig figs, the number of sig figs you have for the two temperatures of the sample.