A 1.10 kg sample of a metal sample absorbs 43.5 kJ of heat, resulting in a temperature rise of 65.0 °C. What is the specific heat capacity of the metal?
Given:
Q = 43.5 KJ = 43500 J
m = 1.10 Kg = 1100 g
ΔT = 65 oC
use:
Q = m*C*ΔT
43500 = 1100.0*C*65.0
C = 0.608 J/g.oC
Answer: 0.608 J/g.oC
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