# 31.5 grams of an unknown substance is heated to 102.4 C and then placed into a calorimeter containing 103.5 grams of water at 24.5 C. If the final temperature reached in the calorimeter is 32.5 C, what is the specific heat of the unknown substance?

Jul 21, 2016

$0.38 C {g}^{-} 1 {\left(\mathrm{de} g r e e C\right)}^{-} 1$

#### Explanation:

Since heat gained by substance equals heat lost by water, and heat transferred is given by
$H = m \cdot c \cdot \Delta \left(\theta\right)$
where, m is the mass of the substance,
c is its specific heat capacity
And $\Delta \left(\theta\right)$ is change in temperature,
$31.5 \cdot c \cdot \left(102.4 - 32.5\right) = 103.5 \cdot 1 \cdot \left(32.5 - 24.5\right)$
Then,
$c = \frac{103.5 \cdot 8}{31.5 \cdot 69.9}$
$= \frac{828}{2201.85}$
$= 0.38$