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