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If a system has 5.00×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (Δ or Δ) of the system?

User Thangaraj
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1 Answer

7 votes
7 votes

Answer:

U=q+w

=5.00×10^2 +5.00×10^2

=2.5×10^5 kJ