Asked by Crimbus
You heat a 5.2 gram lead ball (specific heat capacity=0.128 Joules/g-deg) to 183∘C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity=4.184 J/g-deg) at 22.4 ∘C. What is the final temperature of the water when the lead and water reach thermal equilibrium?
Answers
Answered by
Jaycee
Hot lead loses heat and cold water gains heat.
5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x (T-22.4)
T-22.4 = 106.9/144.3
T-22.4 = 0.7408177408
T = 23.14081774 (round to whatever interval)
5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x (T-22.4)
T-22.4 = 106.9/144.3
T-22.4 = 0.7408177408
T = 23.14081774 (round to whatever interval)
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