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Chemistry question!?
A 30.0-g piece of metal at 203oC is dropped into 200.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate the specific heat of the metal (J/goC). Assume that all of the heat lost by the metal is transferred to the water and no heat is lost to the surroundings.
1 Answer
- ?Lv 77 years ago
Q = mc delta T ... Q = heat energy .. c = specific heat
spec. heat of water = 4.184 j/goC ... water is colder, it gains heat
Q loss = Q gain
(200)(4.184)(29 - 25) = (30)c (203 - 29) solve for c
3347.2 = 5220c
c = 0.64123 J/goC << answer
