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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

Respuesta :

Answer:

Q = 1.5 kJ

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, [tex]T_i=450^{\circ} C[/tex]

Final temperature, [tex]T_f=25^{\circ} C[/tex]

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

[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8\times 0.9\times (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ[/tex]

or

[tex]Q=1.5\ kJ[/tex]

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

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