Thermal+capacity+Question+C

Given information; The heat capacity of a block of meta C= 600 J o C -1

(a) (i)Heat required to raise its temperature by 5 o C can be calculated as C x ΔΘ = 600 x 5 = 3000 J (ii) Heat required to raise its temperature from 30 o C to 50 o C = 600 x 20 =12000 J

(b) C = Q/ΔΘ = 2400/ΔΘ ΔΘ=2400/600=4 The final temperature or the metal block will be 96 o C