# 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)?

Jul 12, 2017

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

$q = m {C}_{\text{s}} \Delta T$

You are given all of the variables needed to solve for $\Delta T$:
$m = 25 \text{g}$
${C}_{s} = \left(0.903 \text{J")/("g" * °"C}\right)$
$q = 2.35 \text{kJ" * (10^3"J")/"kJ" = 2.53*10^3 "J}$

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

$\Delta T = {T}_{f} - {T}_{i}$
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 $\Delta T$ for brevity.