Final Temperature After Phase Change and Transfer of Energy

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Homework Statement

A 0.011 kg cube of ice at 0.0 degrees Celsius is added to 0.450 kg of soup at 80.0 degrees Celsius. Assuming that the soup has the same specific heat capacity as water, find the final temperature of the soup after the ice has melted. (Hint: There is a temperature change after the ice melts.)

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)

Melting point of water: 0.0 degrees Celsius

Latent heat of fusion of water at standard pressure: 3.33 x 10^5 J/kg

answer: approximately 76.2 degrees Celscius

Homework 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

The Attempt at a Solution

... Okay, basically, I tried a bunch of different combinations of Q (Just about everything possible, except for the correct way apparently).

The closest i got to the answer, was 77.11 degrees Celsius, but i am not sure if i arrived at this answer correctly or if it's close enough, because i also came up with 78 degrees Celsius a few times.

1. Substituted values for ice into the equation of Q = m x L
2. Substituted values far ice into the equation of Q = m x Cp x deltaT
3. Substituted values for ice water into the equation of Q = m x Cp x deltaT
4. Added (1.) and (2.) together and set their sum equal to negative (3.)

Could i just get a hint or some other sort of guidance as to how to go about solving this? I have another problem like this, but this one seems to be the simpler of the two.

Just let me know if there is any additional information needed.

 

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First, find the heat needed to convert all ice to water , given by Q=mL. This heat will be taken from the soup. So find the change in temp. of soup by Q=ms(Tf-Ti) , where Tf and Ti are final and initial temperatures. (your Q in this eqn is -ve ), so that final temp is lower than initial temperature.
Now you have two liquids, water at 0 degrees and soup at temp Tf. Just calculate the resulting temperature (T_ff) by -:

$ M_{water} \times s \times (T_{ff} - 0 ) = M_{soup} \times s \times (T_f-T_{ff}) $

 

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thank you for your help.