3990 J of heat are added to a 29.9 g sample
of iron at 28◦C. What is the final temperature of the iron? The specific heat of iron is
0.449 J/g · K.
Answer in units of K.
q = mass Fe x specific heat Fe x (Tfinal-Tinitial)
3990 = 29.9 x 0.449 J/g*c x (Tf - 28)
Solve for Tfinal
Post your work if you get stuck.
Yeah I'm stuck-
3990=29.9x0.449x(Tf-28)
3990=13.4251x(Tf-28)
-13.4251=-13.4252
3976.5749=Tf-28
+28=Tf +28
4004.5749=Tf
To find the final temperature of the iron, we can use the equation:
(q = mcΔT)
where:
q = heat added or removed (in this case, 3990 J)
m = mass of the sample (29.9 g)
c = specific heat capacity of iron (0.449 J/g · K)
ΔT = change in temperature
We need to rearrange the equation to solve for ΔT:
ΔT = q / (mc)
Substituting the given values:
ΔT = 3990 J / (29.9 g * 0.449 J/g · K)
Now, we can calculate the change in temperature:
ΔT = 3990 J / (13.4351 g · K)
ΔT ≈ 297 K
Finally, to find the final temperature of the iron:
Final temperature = initial temperature + ΔT
Final temperature = 28◦C + 297 K
Note that we are adding temperatures in the Kelvin scale, as ΔT is given in Kelvin. To get the answer in Kelvin, we don't need to convert 28◦C to Kelvin since both scales use the same size degree. Therefore, the final temperature of the iron is approximately:
Final temperature ≈ 28 + 297 K
Answer: The final temperature of the iron is approximately 325 K.