# Heat Capacity in Equilibrium Question

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#1
A mass of 0.35kg of ice at −15∘C is lowered into an insulated beaker containing 0.61kg of water at 59∘C. What is the temperature after equilibrium has been reached? Give your answer in ∘C

I've been using the equation m1*c1*ΔT1 = m2*c2*ΔT2 (where m is mass, c is specific heat capacity and ΔT is temperature change) and rearranging it into
Final Temperature = m1*c1*T1 + m2*c2*T2/m1*c1 + m2*c2
but i'm not getting the right answer. I'm not sure where I'm going wrong but some help would be appreciated.
0
2 years ago
#2
5.6 degrees C
0
2 years ago
#3
(Original post by dipsc15)
5.6 degrees C
mcDT+ml+mcDT=mcDT
(0.35*2030*15)+(0.35*3.35x10^5)+ (0.35*4180*T)=(0.61*4180*(59-T))
Ice has to get energy to reach 0 degrees then get energy to melt then get energy to reach equilibrium temp. It gets this energy from the energy released by the 0.61kg water, as it cools down, the 'ice' water heats up, hence getting an answer of 5.6°C
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