Question

What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific heat of Al is 0.900 J/g·C.

Answer 1

LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.