What is the mass of the block of ice?

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SUMMARY

The mass of a block of ice at -13°C that requires a heat transfer of 8.8 x 10^5 J to convert to water at 13°C is calculated using the equation Q=mc(ice)(tf-ti) + mLf + mc(water)(tf-ti). The correct heat of fusion value is 333,000 J/kg, leading to a final mass of 2.11 kg. The initial incorrect calculation of 10.52 kg stemmed from using an incorrect unit for the heat of fusion. Attention to unit accuracy is crucial in thermodynamic calculations.

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  • Understanding of heat transfer principles
  • Familiarity with specific heat capacities of ice and water
  • Knowledge of the heat of fusion for ice
  • Ability to manipulate algebraic equations
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  • Learn about the heat of fusion and its applications in phase changes
  • Practice solving thermodynamics problems involving heat transfer
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chrispsu
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im reviewing for my final in a few days an although this is a basic question i can't work out the correct answer for some reason. and i know i started the other thread below this one but i figured this was a different topic so the heading should be different lol

Homework Statement



A heat transfer of 8.8 x 10^5 J is required to convert a block of ice at -13°C to water at 13°C. What was the mass of the block of ice?

Homework Equations



Q=mc(ice)(tf-ti) + mLf + mc(water)(tf-ti)

The Attempt at a Solution



i tried: 8.8x10^5= m(2220)(13) + m(333) + m(4190)(13)
i come out with the 10.52kg which is not correct.
can someone help pinpoint what i am doing something wrong? thanks!
 
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chrispsu said:
i tried: 8.8x10^5= m(2220)(13) + m(333) + m(4190)(13)
Check those numbers. One of them is wrong - it's in the wrong units.

You need to be very careful with your units!
 
oh ok...the heat of fusion should be 333x10^3...solving it out then i would get 2.11 which is the correct answer. I need to be less careless with my units! lol thanks a lot! :D
 

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