Suppose that 25 g of aluminum is initially at 27.0 C. What is the final temperature of aluminum upon absorbing 2,35 kJ of heat? (#c= 0.903 J/g. ^0C#)?

1 Answer
Jul 12, 2017

This is a specific heat problem that relates heat, mass of a substance, and temperature changing in the following equation:

#q = mC_"s"DeltaT#

You are given all of the variables needed to solve for #DeltaT#:
#m = 25"g"#
#C_s = (0.903"J")/("g" * °"C")#
#q = 2.35"kJ" * (10^3"J")/"kJ" = 2.53*10^3 "J"#

Thus, we're solving an algebraic equation:

#2.35*10^3 "J" = 25"g" * (0.903"J")/("g" *°"C")*DeltaT#
#1.0*10^2 °"C" = DeltaT#

#DeltaT = T_f-T_i#
#1.0*10^2 °"C" = T_f-27.0°"C"#
#T_f approx 131.1°"C"#

If you're following, the reason this is off is because I kept all my significant figures during calculations, but shortened them on #DeltaT# for brevity.