A metal block of mass 11.5 g gives off 360 J of heat while cooling from 99.7 to 32.5 oC. What is the specific heat of the metal in J g-1 oC-1?
q = mass x specific heat x (Tfinal-Tinitial)
To find the specific heat of the metal, we can use the formula:
Q = mcΔT
Where:
Q is the amount of heat energy transferred (in joules),
m is the mass of the metal block (in grams),
c is the specific heat of the metal (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
In this case, we are given:
Q = 360 J,
m = 11.5 g,
ΔT = (Tf - Ti) = (32.5 - 99.7) oC.
Now, we can rearrange the formula to solve for c:
c = Q / (m * ΔT)
Substituting the given values, we have:
c = 360 J / (11.5 g * (32.5 - 99.7) oC)
First, we need to convert the temperature difference from oC to Kelvin:
ΔT = (32.5 - 99.7) oC = -67.2 oC = -67.2 K (since oC and K have the same size unit)
Now we can substitute the values into the equation:
c = 360 J / (11.5 g * -67.2 K)
c = -360 J / (11.5 g * 67.2 K)
c = -5.263 J g^(-1) K^(-1)
Therefore, the specific heat of the metal is approximately -5.263 J g^(-1) K^(-1).