Question

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.?

Answer 1

`40.9kg` lead

Explanation:

The equation you want is

`E=mctheta`

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=mctheta -> m=E/(ctheta)`

We know that `E=93.4J`, `c=0.129J/(gK)` and `theta=40-22.3=17.7K`

(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^oC`, while Kelvin starts at `0` thermal energy, which is about `-273.15^oC`. 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=93.4/(0.129xx17.7)=40.9057kg`