# What is the specific heat of lead if 30.0 g of lead undergoes a 250.0°C change while absorbing 229.5 calories?

Jun 17, 2016

The equation here is $Q = c \cdot m \cdot \Delta T$

#### Explanation:

What we know:
$229.5 = c \cdot 30.0 \cdot 250.0 \to$
$c = \frac{229.5}{30.0 \cdot 250.0} = 0.0306 c a l / g {\cdot}^{o} C$

Note:
It's more usual to use Joules for energy, where $1 c a l \approx 4.16 J$
Also, the unit should be per $k g = 1000 g$ and per Kelvin ($K$), but since it is a temperature change that doesn't make a difference:
$c = 127.3 J / k g \cdot K = 127.3 J \cdot k {g}^{- 1} \cdot {K}^{- 1}$