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what is the temperature of the system (rod + water) at equilibrium assuming no losses to the surroundings?
If a 136.51 g Al rod (c = 0.900 J/g°C) is at 100.0°C and placed into 250.0 g of water at 21.8°C, what is the temperature of the system (rod + water) at equilibrium assuming no losses to the surroundings?
Help please. I'm not even sure where to begin.
2 Answers
- Anonymous1 year agoFavorite Answer
energy given up by the hot Al = energy gained by the cold water
mass of Al * specific heat of Al * (100C - T) = mass of water * specific heat of water * (T - 21.8C)
T is the final temperature, solve for it.
- RealProLv 71 year ago
Well maybe you could begin with google since it's such a common, simple problem?