Can someone help me with this chemistry 🧪 problem?
A 59.4 g sample of metal at 90.81 oC is added to 56.38 g of water that is initially at 16.84 oC. The final temperature of both the water and the metal is 29.36 oC. The specific heat of water is 4.184 J/(goC). Calculate the specific heat of the metal.
- Roger the MoleLv 71 month agoFavorite Answer
(4.184 J/g·°C) x (56.38 g) x (29.36 - 16.84)°C =
2953.39188 J gained by the water (and lost by the metal)
(2953.39188 J) / (59.4 g) / (90.81 - 29.36)°C = 0.809 J/g·°C
[Although it is difficult to think of an actual metal with a specific heat like that.]