(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

find the final temperature of A 150g (0,15kg) cube of ice at 0.0 degrees Celsius is added to 300g(0.3 kg) of water at 50.0 degrees Celsius.

Specific heat capacity of ice: 2.09 x 10^3 J/(kg x degrees Celsius)

Specific heat capacity of water: 4.186 x 10^3 J/(kg x degrees Celsius)

2. Relevant equations

Q = m x L

energy transferred as heat during a phase change = mass x latent heat

Q = m Cp deltaT

energy transferred as heat = mass x specific heat capacity x change in temperature

3. The attempt at a solution

I wouldnt had problem to solve if it wasnt for the phase change

For the water q=mcdeltaT

(0,3kg)*(4190j/kg*k)(T-50C)

If it was no phase its the same for the other liquid and then just set them equal to = 0 and find the T but no with the Phase it will be a bit different but how ?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Final Temperature After Phase Change

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