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.?

1 Answer
Mar 2, 2017

Answer:

#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#