A 4.70 g nugget of pure gold absorbed 283 J of heat. What was the final temperature of the gold if the initial temperature was 25.0°C? The specific heat of gold is 0.129 J/(g·°C)
First right gets 10 points
First right gets 10 points
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Q = mc(delta t)
delta t = Q/mc
delta t = 283 J /(4.70 g)(0.129 J/g C)
delta t = 467 C
If it started at 25 C and went up by 467 C, the final temp would be 492 C
delta t = Q/mc
delta t = 283 J /(4.70 g)(0.129 J/g C)
delta t = 467 C
If it started at 25 C and went up by 467 C, the final temp would be 492 C