# Given the specific heat of lead is 0.129 J/g*K and that it takes 93.4 J of energy to heat a sample of lead from 22.3°C to 40.°C, how do you find the mass of the lead.?

Mar 2, 2017

$40.9 k g$ lead

#### Explanation:

The equation you want is

$E = m c \theta$

where $E$ is energy, $m$ is mass, $c$ is specific heat capacity (SHC) and $\theta$ is the change in temperature (note: change).

In the question, we want to find mass, so we rearrange to make mass the subject:

$E = m c \theta \to m = \frac{E}{c \theta}$

We know that $E = 93.4 J$, $c = 0.129 \frac{J}{g K}$ and $\theta = 40 - 22.3 = 17.7 K$

(We can easily convert between $K$ and Celsius because they are on the same scale, but have different starting points. Celsius starts from water freezing at ${0}^{o} C$, while Kelvin starts at $0$ thermal energy, which is about $- {273.15}^{o} C$. Calculating a change in Celsius gives the same change in Kelvin, because you can ignore the starting points of the scales.)

Now we have

$m = \frac{93.4}{0.129 \times 17.7} = 40.9057 k g$