Question

Given the specific heat for aluminum is 0.900 J/g.°C, how much heat is released when a 3.8 g sample

of Al cools from 450.0°C to 25°C?
A. 1.5 kJ
B. 54 J
C. 60J
D. 1.7 kJ
E. 86 J

Answer 1

Q = 1.5 kJ

Step-by-step explanation:

It is given that,

The specific heat for aluminum is 0.900 J/g°C

Mass of sample, m = 3.8 g

Initial temperature,
`T_i=450^(\circ) C`

Final temperature,
`T_f=25^(\circ) C`

We need to find the heat released. The amount of heat released is given by the formula:

`Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8* 0.9* (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ`

or

`Q=1.5\ kJ`

So, the correct option is (A) i.e. 1.5 kJ.