A 100 g ice cube at -10.0 °C is placed in an aluminum cup whose initial temperat

Question

A 100 g ice cube at -10.0 °C is placed in an aluminum cup whose initial temperature is 70.0 °C. The system comes to an equilibrium temperature of 20.0 °C. What is the mass of the cup? Assume all the heat transfers take place between the ice/water and aluminum cup. The specific heat of water and ice are 4186 J/kg/K and 2090 J/kg/K The latent heat of fusion for water/ice is 3.33 × 105 J/kg. (answer: mcup = 970 g) Hint: There are 4 heat energy exchanges before the system reaches thermal equilibrium

Explanation / Answer

here,

mass of ice , mi = 0.1 kg

initial temprature of ice , Ti = - 10 degree C

initial temprature of alumunium , Ta = 70 degree C

final temprature , Tf = 20 degree C

let the mass of alumunium be ma

heat lost by alumunium = heat gained by ice

ma * Cal * ( Ta - Tf) = mi * ( Ci * ( Ti - 0) + Li + Cw * ( Tf - 0))

ma * 910 * ( 70 - 20) = 0.1 * ( 2090 * (10 - 0) + 33300 + 4186 * (20 - 0))

ma = 0.303 kg

ma = 303 g

the mass of the cup is 303 g

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