If a 30 gram sample of Copper shots (specific heat = 0.385 J/gºC) changed from 27ºC to 90ºC, how much heat was involved?

728 J

-728 J

1,351 J

-1,351 J

728J

Explanation:

The following data were obtained from the question:

Mass (M) = 30g.

specific heat capacity (C) = 0.385 J/gºC.

Initial temperature (T1) = 27ºC

Final temperature (T2) = 90ºC

Change in temperature (ΔT) = T2 – T1 = 90°C – 27 = 63°C

Heat (Q) =...?

The heat involved can be obtained as illustrated below:

Q = MCΔT

Q = 30 x 0.385 x 63

Q = 728J

Therefore the heat involved is 728J