Calc 320g H2O Melting Ice @ 68°C

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    Ice Melting
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Homework Help Overview

The problem involves calculating the amount of ice that can be melted by a specific mass of water at a given temperature. The context is centered around thermodynamics, specifically the concepts of heat transfer and phase changes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of heat transfer equations, including the use of latent heat and specific heat in the calculations. There is a focus on determining the energy change as water cools and how that relates to melting ice.

Discussion Status

The discussion includes attempts to clarify the calculations involved and confirm the correctness of the approach taken. Some participants provide references for further reading, indicating a collaborative effort to deepen understanding.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available and the methods that can be employed. There is an assumption that the latent heat values and specific heat capacities are known and applicable.

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How much ice could be melted by 320. grams of water at 68.0 degrees C?
Would I us mL + mcT=0 for the equation, and do the latent heat of ice for the first part and water for the second?
 
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One must determine the energy (heat) from the mass of water decreasing from 68°C to 0°C. Then determine what mass of ice would be transformed to water at 0°C.

Assuming that L in your equation is latent heat, and by T you mean change in temperature, you are correct.

The heat from the liquid, Q, is just the change in energy in the liquid, i.e. Q = \Delta\,H = m\,c\,\Delta\,T, where c is specific heat.
 
So it would be m(333.7kJ/kg)+(.320kg)(4.186kJ/kg*K)(-68.0K)=.273kg of ice Is this correct?
 

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