b) A 20kg mass of Aluminum of specific heat Capacity OF 900kg /k was heated from 2k to 12k. Determine the
amount of heat Energy required
Q=cmΔT=900Jkg×K×20 kg×(12 K−2 K)=180000 J=180 kJQ=cm\Delta{T}=900\frac{J}{kg\times{K}}\times20\ kg\times(12\ K-2\ K)=180000\ J=180\ kJQ=cmΔT=900kg×KJ×20 kg×(12 K−2 K)=180000 J=180 kJ
Answer: 180 kJ
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