The specific heat of iron is 0.449 J/(g . oC) and the specific heat of water is

Question

The specific heat of iron is 0.449 J/(g . oC) and the specific heat of water is 4.184 J/(g . oC). A piece of iron was heated to 98.1oC and dropped into a constant pressure calorimeter containing 262 grams of water at 34.9oC. The final temperature of the water and iron was 48.4oC. Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass, in g, of the piece of iron? A) 7.64 g B) 159 g C) 1.51 x 10-03 g D) 683 g E) 6.31 x 10-03 g F) 663 g G) 714 g I was using q=m *s* delta t but was getting big number like 2441.11 g so i'm not sure what i am doing wrong

Explanation / Answer

the iron and water end up at the same temp, so iron loses heat and water gains heat iron's loss = water's gain m cp delta t = m cp delta t m iron = (m water) (cp w) (delta T water)/[(cp iron)(delta T iron)] m iron = 262*4.184*(48.4-34.9)/[0.449*(98.1-48.4)] = 663g Even though the iron drops in temperature more degrees and is more massive, the cp of water is ten time bigger and that outweighs the other factors.

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