What is the specific heat capacity of lead if a 18.9 g sample increased 12.9 C upon the addition of 31.5 J? The units are J/(gK)
q = mass Pb x specific heat Pb x delta T.
So is this how i do this
31.5J= (18.9)(12.9)
Yes, except you didn't put in an unknown. The way you have it written 31.5 IS NOT equal to 18.9 x 12.9.
What you want is 31.5 = (18.9)(X)(12.9) and solve for X.
I assume you just made a typo and omitted the unknown of X.
Yeah i forgot the x. Thank you
To calculate the specific heat capacity of lead, you can use the formula:
q = mcΔT
Where:
q is the heat energy transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in J/(gK)), and
ΔT is the change in temperature (in degrees Celsius or Kelvin).
In this case, you are given:
mass (m) = 18.9 g
change in temperature (ΔT) = 12.9 °C
heat energy transferred (q) = 31.5 J
To calculate the specific heat capacity (c), you need to rearrange the formula:
c = q / (m * ΔT)
Plug in the given values:
c = 31.5 J / (18.9 g * 12.9 °C)
To ensure consistent units, convert the change in temperature from Celsius to Kelvin by adding 273.15:
c = 31.5 J / (18.9 g * (12.9 °C + 273.15))
Now you can solve for c.