Final temperature of two mixed substances (simply enthelpy math)

AI Thread Summary
The discussion revolves around calculating the final temperature of a mixture of copper and water using the principle of heat transfer. The equation used is Q = m * ΔT * c, where heat lost by copper equals heat gained by water. The initial calculations led to an incorrect final temperature of 19.725°C, which is lower than both initial temperatures. The error identified is the assumption that the final temperature of copper could exceed its initial temperature of 100°C. The correct final temperature, as indicated, should be 29.6°C.
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Homework Statement



A 100-g sample of Cu(s), initially at 100oC, is added to 140 g of water, initially at 25oC. What is the final temperature of the mixture?

Specific heat
(in J g-1 oC-1)
Cu(s) 0.385
H2O(l ) 4.184

Homework Equations



Q = m * ΔT * c

heat lost = heat gained
= m2 * ΔT2 * c2

The Attempt at a Solution



100 * (T2 - 100) * 0.385 = 140 * (T2 - 25) * 4.185
38.5* (T2 - 100) = 585.9 * (T2 - 25)
38.5T2 - 3850 = 585.9T2 - 14647.5
T2 = 19.725°CBut the temperature can't be lower than both their starting temperatures, what did I do wrong? The answer is supposed to be 29.6
 
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From the way you have written your equation, you are assuming that the final temperature T2 of the copper is greater than 100 C.
 
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