If you are given the temperature of a drink (iced tea) andgiven the mass of ice

Question

If you are given the temperature of a drink (iced tea) andgiven the mass of ice cubes that are in the drink and asked what isthe final entropy change of the ice cubes how would you findit? I thought since no further information was given on the icedtea I would use the temperature given as the final temperature forthe ice cubes. So what I did was multiply the mass by the heat offusion and added that value to the heat for the ice which I foundby multiplying the mass, specific heat of ice, and delta T. Butthat was incorrect. I even tried adding the heat (Q) of the water (because the ice melted) at the same temperaturebut I figured it was incorrect because if the ice cubesmelted wouldn't the temperature change??? Here are hypothetical numbers: Temp of ice tea: 276 K Mass of ice: 0.12 kg I did (0.12 kg)*(333.5) + (0.15 kg)*(2.05)*(276) WRONG! I'm lost? If you are given the temperature of a drink (iced tea) andgiven the mass of ice cubes that are in the drink and asked what isthe final entropy change of the ice cubes how would you findit? I thought since no further information was given on the icedtea I would use the temperature given as the final temperature forthe ice cubes. So what I did was multiply the mass by the heat offusion and added that value to the heat for the ice which I foundby multiplying the mass, specific heat of ice, and delta T. Butthat was incorrect. I even tried adding the heat (Q) of the water (because the ice melted) at the same temperaturebut I figured it was incorrect because if the ice cubesmelted wouldn't the temperature change??? Here are hypothetical numbers: Temp of ice tea: 276 K Mass of ice: 0.12 kg I did (0.12 kg)*(333.5) + (0.15 kg)*(2.05)*(276) WRONG! I'm lost?

Explanation / Answer

energy necessary to melt ice cubes = Q =mLf     where m is mass of ice cube, Lis latentheat of ice      T = 276K change in entropy = S = Q/T plug values and do caliculations

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