Ice at 00C  is placed in a Styrofoam cup containing 0.32 kg of lemonade at 270C . The specific heat capacity of the lemonade is virtually the same as that of water, c = 4186 J/kg 0C. After the ice and lemonade reached the equilibrium temperature, some ice still remains. The latent heat of fusion for water is Lf = 3.35 x105 J/kg. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. Determine the mass of ice that has melted.

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A:

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A: Basic Details The heat transfers from the water at high temperature to ice at low temperature. The…

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A: Specific heat of water= 4186 J/kg.K

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A: Write the given values of this problem. m1=20 g or 0.02 kgT1=2°C or 275 Km2=300 g or 0.3 kgT2=170°C…

Q: What is the temperature of the container and water once they reach thermal equilibrium

A: Complete explanation and final answer is in next step.

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A: Recall Q=mc∆T Also, principle of calorimetry. Heat lost by hotter body=heat gained by colder body

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A: The solution for this question has been given below..

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A:

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Q: You are heating 310 mL of water to make tea; however, you forget the pot on the stove, and all of…

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Q: An insulated Thermos contains 150 cm³ of hot coffee at 90.0°C. You put in a 11.0 g ice cube at its…

A:

Q: A 34.0-g ice cube floats in 234 g of water in a 100-g copper cup; all are at a temperature of 0°C. A…

A: Given mass of ice cube mi=34gmass of water mw=234g

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A: SOlution: Given that m = 10 kg L =  6.44×104 J/kg

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A: