# 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?

Jun 19, 2017

I get ${26}^{\circ} \text{C}$. 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 ${4}^{\circ} \text{C}$.

This illustrates that heat capacity ($m \times {C}_{P}$, in $\text{J/"^@ "C}$) 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 $q$:

$q = m {C}_{P} \Delta T$,

where:

• $m$ is the mass in $\text{g}$.
• ${C}_{P}$ is the specific heat capacity in $\text{0.49 J/g"^@ "C}$ at constant pressure, the amount of heat required to increase the temperature of one gram of substance by ${1}^{\circ} \text{C}$.
• $\Delta T = {T}_{2} - {T}_{1}$ is the change in temperature in $\text{^@ "C}$.

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

$D = \frac{m}{V}$

You gave the volume as

${\text{25 cm" xx "45 cm" xx "2.5 cm" = "2812.5 cm}}^{3}$

So, the mass is given by

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

$=$ $\text{21937.5 g}$

You also gave that the heat that went in was

$q = + \text{3.8 kJ" = +"38000 J}$

(which you should remember to make sure is in the same energy unit, $\text{J}$, as in ${C}_{P}$, and not $\text{kJ}$!)

and the specific heat capacity as

${C}_{P} = \text{0.49 J/g"^@ "C}$.

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 ${T}_{1} = {22}^{\circ} \text{C}$:

$q = m {C}_{P} \Delta T$

"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 $m {C}_{P}$ to get:

$\frac{38000}{10749.375} \text{^@ "C" = 3.535^@ "C" = T_2 - 22^@ "C}$

Therefore, the new temperature is:

$\textcolor{b l u e}{{T}_{2}} = 22 + 3.535 = {25.535}^{\circ} \text{C}$

$= \textcolor{b l u e}{{26}^{\circ} \text{C}}$ to two sig figs.