# Solid iron a specific heat of 0.444 J/g°C. If 75 calories of energy were removed from a piece of iron, and the temperature of the iron went from 21.4°C to -25.2°C, what was the mass of the iron?

##### 1 Answer

Jul 21, 2016

When you look at the big picture, what you have is heat flowing out of solid iron and cooling it down.

So, you have to use the **heat flow equation**:

#\mathbf(q = msDeltaT)# where:

#q# is theheat flowin#"J"# .#m# is themassof the object in#"g"# .#s# is thespecific heat capacityof the object in#"J/g"cdot""^@ "C"# .#DeltaT# is thechange in temperatureof the iron in#""^@ "C"# .

First, convert the heat from *out* is *negatively-signed* by convention.

#-75# #cancel"cal" xx "4.184 J"/cancel"cal" = -"313.8 J"#

Now what you have is:

#color(blue)(m_"Fe") = q_"Fe"/(s_"Fe"DeltaT_"Fe")#

#= q_"Fe"/(s_"Fe"[T_f^"Fe" - T_i^"Fe"])#

#= (-313.8 cancel("J"))/((0.444 cancel("J")"/g"cdotcancel(""^@ "C"))[(-25.2cancel(""^@ "C")) - (21.4cancel(""^@ "C"))])#

#=# #color(blue)("15.2 g Fe")#