# Chem Help????

## An alloy has a specific heat of 1.77 cal/g degrees C and is at 25 degrees C. When it's dropped into warm water, the alloy absorbs 229 cal of heat. If the final temperature of the alloy is 37 degrees C, what is it's mass? Help! I don't understand ANYTHING in this subject. Explanation and the answer??

Apr 3, 2018

$\approx$ 10.8 grams

#### Explanation:

This is a common question regarding heat absorption of a material. Seems like we use Calories as an energy unit, which is a bit unconventional, but nontheless here is how to do it:

Using the equation $Q = m c \Delta T$

$Q$= energy input (Calories in this case, otherwise joules)
$m$=mass of the object, in grams, because $c$ says calories per gram.
$c$- the specific heat capacity of the object, which is the energy input needed to raise a specific material's temperature by 1 $K$.
$\Delta T$- the temperature change of the material in Kelvin (but again, a change in temp of 1 Kelvin is equal to a change in temp of 1 Celcius.)

Now from interpreting the question, it seems the alloy was 25 degrees C at first but heated up to 37 degrees after being put in hot water-so it heated up by 12 degrees Celsius (or Kelvin, as the scales are related). So $\Delta T = 12$

In the question, we are given $c = 1.77$ cal/g and $Q = 229$ cal. The only thing missing is thus the mass!

Plugging everything in:
$Q = m c \Delta T$

$229 = m \times 1.77 \times 12$

$m = \frac{229}{1.77 \times 12}$

$m = 10.78 g \approx 10.8 g$