# 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?

##### 1 Answer

I get

This illustrates that heat capacity (

This overall will be using the equation for **heat flow**

#q = mC_PDeltaT# ,where:

#m# is themassin#"g"# .#C_P# is thespecific heat capacityin#"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 thechange in temperaturein#""^@ "C"# .

The extra step here is to find the mass using the steel bar's volume **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,

#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

#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.