Specific heat problem finding final temp with calorimeter system

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SUMMARY

The forum discussion centers on calculating the final temperature of a calorimeter system involving a 34 g block of ice at -75°C, 562 g of water, and an 80 g copper calorimeter at 22°C. The specific heat capacities used are 387 J/kg·°C for copper, 2090 J/kg·°C for ice, and 4186 J/kg·°C for water, with the latent heat of fusion for ice being 3.33 x 10^5 J/kg. The correct approach involves applying the heat transfer equation Q = mcΔT, ensuring proper sign conventions for heat flow, leading to the accurate calculation of the final temperature.

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Homework Statement



A 34 g block of ice is cooled to -75°C. It is added to 562 g of water in an 80 g copper calorimeter at a temperature of 22°C. Find the final temperature. (The specific heat of copper is 387 J/kg ·°C and the specific heat of ice is 2090 J/kg ·°C. The latent heat of fusion of water 3.33 x 10^5 J/kg and the specific heat of water is 4186 J/kg ·°C.) Answer in units of °C.



Homework Equations



1,000 g= 1 kg

Q= mcdeltaT

Qice + Qlost H20 + Qcopper containter =0

mcdeltaT of ice + mL of ice + Qlost H20 + Qcopper container =0



The Attempt at a Solution



I tried the last equation.

.034(2090) (75) + (.034) (3.33e-9) + (.562)(4186)(Tf-22) + (.08)(387)(Ttf-22)=0

and i got 134.28 C which was wrong. Help!
 
Physics news on Phys.org
Check the signs of the dT
which have heat flowing in and which out
 
i actually just got the answer, but thanks!
 

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