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3 November, 07:21

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

+1
Answers (2)
  1. 3 November, 07:34
    0
    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
  2. 3 November, 08:25
    0
    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
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