Answer
Therefore 612 J of heat are released.
Work Step by Step
We will use the equation $q=mc\Delta t$
q = heat transferred
m = mass
c = specific heat
$\Delta t $ = change in temperature
plug in the values.
q = (25.7) (0.44) (-53)
*** solve using a calculator
q= -612 J
Therefore the heat transferred is 612 J and since the sign is negative it means that the heat was released. Therefore 612 J of heat are released.
