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Mkey [24]
3 years ago
13

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 fr

om 25.0°C to 76.0°C?
Chemistry
1 answer:
faust18 [17]3 years ago
4 0

Answer:

quantity of heat=mc*theta

=25*0.930(76-25)

=25*0.930*51

=1185.75J

=11.9kJ

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