A 4.0 gram sample of iron was heated from 0 degrees celsius to 20 degrees celsius, and was found to have absorbed 35.2 Joules of heat. Calculate the specific heat of this piece of iron.
Q=mc∆t
find c.
http://www.ausetute.com.au/heatcapa.html
Formula for specific heat:q/(m * delta t)
Input given data: 35.2 J/(4.0 g * 20 c)
Solve: 0.44 J/(g*c)
To calculate the specific heat of the piece of iron, we need to use the formula:
Q = mcΔT
Where:
Q is the heat absorbed (in Joules)
m is the mass of the substance (in grams)
c is the specific heat (in J/g°C)
ΔT is the change in temperature (in °C)
In this case, we know the following values:
Q = 35.2 J
m = 4.0 g
ΔT = 20°C - 0°C = 20°C
So, we can rearrange the formula to solve for c:
c = Q / (m * ΔT)
Substituting the known values:
c = 35.2 J / (4.0 g * 20°C)
c = 0.44 J/g°C
Therefore, the specific heat of this piece of iron is 0.44 J/g°C.