A 45.0-gram sample of copper metal was heated from 20.0°C to 100.0°C. Calculate the heat absorbed, in kJ, by the metal.

1.386 KJ

Explanation:

From the question given above, the following data were obtained:

Mass (M) of copper = 45 g

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Heat absorbed (Q) =..?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 100 – 20

ΔT = 80 °C

Next, we shall determine the heat absorbed by the sample of copper as follow:

Mass (M) of copper = 45 g

Change in temperature (ΔT) = 80 °C

Specific heat capacity (C) of copper = 0.385 J/gºC

Heat absorbed (Q) =..?

Q = MCΔT

Q = 45 × 0.385 × 80

Q = 1386 J

Finally, we shall convert 1386 J to KJ. This can be obtained as follow:

1000 J = 1 KJ

Therefore,

1386 J = 1386 J × 1 KJ /1000 J

1386 J = 1.386 KJ

Thus, the heat absorbed by the sample of the sample of copper is 1.386 KJ.

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.