Final temperature of two mixed substances (simply enthelpy math)

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SUMMARY

The final temperature of a mixture consisting of 100 g of copper (Cu) at 100°C and 140 g of water (H2O) at 25°C is calculated using the principle of heat transfer. The specific heat capacities are 0.385 J/g°C for copper and 4.184 J/g°C for water. The equation Q = m * ΔT * c is applied, leading to the equation 100 * (T2 - 100) * 0.385 = 140 * (T2 - 25) * 4.184. The correct final temperature is 29.6°C, indicating an error in the initial assumption that T2 could exceed 100°C.

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  • Understanding of specific heat capacity
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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
 
Last edited:
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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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