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

A = 728J

Explanation:

Mass of substance = 30g

Specific capacity of copper = 0.385J/g°C

Initial temperature (T1) = 27°C

Final Temperature (T2) = 90°C

Heat energy (Q) = ?

Heat Energy (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature of the substance = T2 - T1

Q = mc∇T

Q = mc * (T2 - T1)

Q = 30 * 0.385 * (90 - 27)

Q = 11.55 * 63

Q = 727.65J

The heat energy required to raise 30g of copper sample from 27°C to 90° is 728J