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?

Question

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?

Jaycee · Answer

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)