Half a litre of water (mass=0.500kg) is contained in a 0.150kgglass beaker at 20
ID: 1756616 • Letter: H
Question
Half a litre of water (mass=0.500kg) is contained in a 0.150kgglass beaker at 20 degrees celcius. An ice mass of 0.045kg isadded. The final temperature of the system is 12 degreescelcius. You may assume that no energy exchange occursbetween the system and the surroundings. a) How much energy is lost by the water that was originally inthe glass? b) How much energy is required to melt the ice? c) What was the original temperature of the ice? Half a litre of water (mass=0.500kg) is contained in a 0.150kgglass beaker at 20 degrees celcius. An ice mass of 0.045kg isadded. The final temperature of the system is 12 degreescelcius. You may assume that no energy exchange occursbetween the system and the surroundings. a) How much energy is lost by the water that was originally inthe glass? b) How much energy is required to melt the ice? c) What was the original temperature of the ice?Explanation / Answer
mass of water m = 0.5 kg mass of glass beaker m ' = 0.150 kg mass of ice M = 0.045 kg Initial temperature of water t = 20 oC final temp of the system T = 12 o C initial temp of ice t ' = ? (a).energy is lost by the water that was originally in theglass Q = mc dt where c = specific heat of water = 4186 J / kg oC dt =temp difference = t - T = 8 oC substitue values we get Q = 16744 J (b). Energy required to melt the ice Q " = ML where L = latent heat of water = 334880 J /kg substitue values we get Q " = 15069.6 J (b). Heat lost by water + glass beaker = heatgain by ice Q + m ' c ' dt = M C dt ' + Q " + Mc dt " where c ' = specific heat of glass = 500 J /kg oC C = specific heat of ice = 2100 J / kg oC dt ' =t ' dt " =12 oC substitue values we have 16744 + 83 = 94.5 t ' + 15069.6 +2260.44 from this initial temperature of the ice t ' = -5.323oCRelated Questions
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