Question

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`)?

Answer 1

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.