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# Specific Heat Capacity Help!!?

11.0 J of heat is added to 10.00 g of gold, which is initially at 25°C.

The heat capacity of gold is 25.41 J/(mol • °C).

What would be the final temperature of the gold?

after I convert to moles I get 0.0508

### 1 Answer

Relevance

- Roger the MoleLv 73 weeks ago
(11.0 J) / (25.41 J/(mol • °C)) / (10.00 g Au / (196.966569 g Au/mol)) =

8.53°C change

25°C + 8.53°C = 33.5°C

[Your number of moles is correct to 3 significant digits.]

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