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How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 25.0°C to 76.0°C?

Respuesta :

apokijoy3

Answer:

quantity of heat=mc*theta

=25*0.930(76-25)

=25*0.930*51

=1185.75J

=11.9kJ

afreeze2323

Answer:

1.19kJ

Explanation:

The heat can be calculated using the specific heat capacity, mass, and the change in temperature.

q=mxcxΔT=(25.0g)×(0.930Jg⋅∘C)×(76.0−25.0∘C)=1186J

Finally, the heat can be converted from J into kJ using the metric conversion factor.

1186J×1kJ1000J=1.19kJ

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