What was the initial temperature of the copper?

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Homework Help Overview

The problem involves a calorimetry scenario where aluminum and copper are mixed with ethyl alcohol, and the goal is to determine the initial temperature of the copper based on the final equilibrium temperature.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using calorimetry equations to relate heat lost and gained, with some expressing difficulty in applying these concepts after a long time. There are attempts to set up equations based on mass, specific heat capacity, and temperature changes.

Discussion Status

The discussion is ongoing, with participants sharing their attempts to solve the problem and questioning the setup of the equations. Some guidance has been offered regarding the calorimetry approach, but there is no clear consensus on the correct method or solution yet.

Contextual Notes

One participant notes the absence of water in the problem, which may indicate a misunderstanding or misinterpretation of the scenario. There are also indications of confusion regarding the specific heat capacities and mass values used in the calculations.

bigboss
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Homework Statement



8.0 g of aluminum at 200 C and 22 g of copper are dropped into 55 cm^3 of ethyl alcohol at 15 C. The temperature quickly comes to 28 C. What was the initial temperature of the copper?

Homework Equations


The Attempt at a Solution



tried using calorimetry equations, not getting anywhere.
 
Last edited:
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Have to give the equations i haven done SHC for 3years although i don't think that that question is hard.
 
heat lost = heat gained
or m1 * s1 * t1 + m2 * s2 * t2 + m3 * L = m1 * s1 * tf + m2 * s2 * tf + m3 * s3 * tf
solving the above equation for t2 we get
t2 = (m1 * s1/m2 * s2) * (tf - t1) + (m3/m2 * s2) * (s3 * tf - L)
m1 = 11 g = 11 * 10^-3 kg
s1 = 900 J/kg/K
t1 = 200 oC = (200 + 273) K = 473 K
m2 = 21 g = 21 * 10^-3 kg
s2 = 387 J/kg/K
t2 = the initial temperature of copper
m3 = ρ * V = 790 * 46 * (10^-2)^3 = 36.34 * 10^-3 kg
L is the latent heat of fusion of ethyl alcohol and is equal to 1.08 * 10^5 J/kg
tf = 22 oC = (22 + 273) K = 295 Knot working.
 
Last edited:
I do not see any water in the problem.

ehild
 
thanks for the notice, edited it, and stil not getting the correct answer
 

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