Question

Suppose you have a `25xx45xx2.5` `"cm"` steel bar whose density is `"7.8 g/cm"^3` at `22^@ "C"`. If `"3.8 kJ"` was imparted into it by solar energy, and the bar's specific heat capacity is `"0.49 J/g"^@ "C"`, to what temperature does it rise?

Answer 1

I get `26^@ "C"`. Since the mass of the steel bar was large, even though it absorbed a lot of solar energy (heat), it didn't increase much in temperature, only `4^@ "C"`.

This illustrates that heat capacity (`m xx C_P`, in `"J/"^@ "C"`) is an extensive property, whereas specific heat capacity, `C_P`, is an intensive property. The specific heat capacity is independent of how large the object is.

---

This overall will be using the equation for heat flow `q`:

`q = mC_PDeltaT`,

where:

• `m` is the mass in `"g"`.
• `C_P` is the specific heat capacity in `"0.49 J/g"^@ "C"` at constant pressure, the amount of heat required to increase the temperature of one gram of substance by `1^@ "C"`.
• `DeltaT = T_2 - T_1` is the change in temperature in `""^@ "C"`.

The extra step here is to find the mass using the steel bar's volume `V` in `"cm"^3` and density `D` in `"g/cm"^3`. The density is given by:

`D = m/V`

You gave the volume as

`"25 cm" xx "45 cm" xx "2.5 cm" = "2812.5 cm"^3`

So, the mass is given by

`m = DV = "7.8 g"/cancel("cm"^3) xx 2812.5 cancel("cm"^3)`

`=` `"21937.5 g"`

You also gave that the heat that went in was

`q = +"3.8 kJ" = +"38000 J"`

(which you should remember to make sure is in the same energy unit, `"J"`, as in `C_P`, and not `"kJ"`!)

and the specific heat capacity as

`C_P = "0.49 J/g"^@ "C"`.

As a result, we now have enough information to determine the final temperature, `T_2`, that the bar is now at. Using the first equation and the given `T_1 = 22^@ "C"`:

`q = mC_PDeltaT`

`"38000 J" = (21937.5 cancel"g")("0.49 J/"cancel"g"^@ "C")(T_2 - 22^@ "C")`

`"38000 J" = overbrace(("10749.375 J/"^@ "C"))^("Heat Capacity, " mC_P)(T_2 - 22^@ "C")`

Divide over the heat capacity `mC_P` to get:

`38000/10749.375 ""^@ "C" = 3.535^@ "C" = T_2 - 22^@ "C"`

Therefore, the new temperature is:

`color(blue)(T_2) = 22 + 3.535 = 25.535^@ "C"`

`= color(blue)(26^@ "C")` to two sig figs.