Suppose you have a $25 \times 45 \times 2.5$ $25xx45xx2.5$ $cm$ $"cm"$ steel bar whose density is ${7.8 g/cm}^{3}$ $"7.8 g/cm"^3$ at $22^{\circ} C$ $22^@ "C"$ . If $3.8 kJ$ $"3.8 kJ"$ was imparted into it by solar energy, and the bar's specific heat capacity is ${0.49 J/g}^{\circ} C$ $"0.49 J/g"^@ "C"$ , to what temperature does it rise?

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Suppose you have a $25 \times 45 \times 2.5$ $25xx45xx2.5$ $cm$ $"cm"$ steel bar whose density is ${7.8 g/cm}^{3}$ $"7.8 g/cm"^3$ at $22^{\circ} C$ $22^@ "C"$ . If $3.8 kJ$ $"3.8 kJ"$ was imparted into it by solar energy, and the bar's specific heat capacity is ${0.49 J/g}^{\circ} C$ $"0.49 J/g"^@ "C"$ , to what temperature does it rise?

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Answer 1 · 2017-06-19T23:17:22

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

This illustrates that heat capacity ( $m \times C_{P}$ $m xx C_P$ , in ${J/}^{\circ} C$ $"J/"^@ "C"$ ) is an extensive property, whereas specific heat capacity, $C_{P}$ $C_P$ , 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 $q$ $q$ :

$q = mC_PDeltaT$ ,

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.

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