How much heat is absorbed when a block of copper of mass 0.05kg and specific heat capacity 390Jkg-1K-1 is heated from 20oC to 70oC?
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Correct Answer: Option B
Explanation:
Q = mc\(\theta\)
= 0.05 x 390 x (70 - 20)
= 975J
Q = mc\(\theta\)
= 0.05 x 390 x (70 - 20)
= 975J