The specific thermal capacity of Al is c=0.902 Jg⋅°C;c = 0.902~\frac{\text{J}}{\text{g}\cdot\degree\text{C}};c=0.902 g⋅°CJ;
ΔQ=cmΔt=0.902 Jg⋅°C ∗ 10.9 g ∗ (36.7−20.0) °C≈164.19 J.{\Delta}Q = cm{\Delta}t = 0.902~\frac{\text{J}}{\text{g}\cdot\degree\text{C}}~*~10.9~\text{g}~*~(36.7-20.0)~\degree\text{C} \approx 164.19~\text{J}.ΔQ=cmΔt=0.902 g⋅°CJ ∗ 10.9 g ∗ (36.7−20.0) °C≈164.19 J.
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