# What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3 calories of heat and the temperature rises 15.0°C?

Nov 25, 2017

See below.

#### Explanation:

The equation used to solve for specific heat capacity is this:

$q = s \cdot m \cdot \Delta T$ where q heat absorbed or lost, s is the specific heat capacity, m is the mass, and $\Delta T$ is the change in temperature $\text{final temperature"-"initial temperature}$

From what the problem gave us, we can plug in the numbers:

$47.3 = s \cdot 55.00 \cdot 15.0$

Simplify:

$47.3 = s \cdot 825$

Divide by 825 on both sides:

$0.0573 = s$

The units of $s$ will be in "cal"/(g*C for this problem, so $s = 0.0573 \frac{\text{cal}}{g \cdot C}$