A. Find the amount of heat that must be extracted from 1kgof steam at a temperat

Question

A. Find the amount of heat that must be extracted from 1kgof steam at a temperature of 110°C in order to convert all ofit to ice at 0°C. What speed would this correspond to ifthis amount of thermal energy were equal to translational kineticenergy for the given mass?

B. A 50g ice cube is added to 110g of water in a 62galuminum cup. The water and cup initially have a temperatureof 23°C. Find the initial temperature of the ice cubesuch that all of the water is just turned to ice. Use thefollowing specific heats: water = 4186 J/(kg·K), ice = 2090J/(kg·K), and aluminum = 900 J/(kg·K). Thelatent heat of fusion for water is 33.5e4 J/kg

Explanation / Answer

mass m = 1kg Initial temp of stream t = 110 oC temperature of ice t ' = 0 oC Required amount of heat energy Q = mC dt + mL + mc dt '+ mL '                                                      = m[ C dt + L + c dt ' + L '] where C = specific heat of steam = 0.5 cal / g            L= latent heat of vaporisation = 540 cal / g            c = specific heat of water = 1 cal / g oC            L ' = latent heat of fusion = 80 cal / g            dt = 110 - 100 = 10 oC            dt ' = 100- 0 = 100 oC plug the values we get Q = 1 * [ 5 + 540 + 100+ 80]                                       = 725 cal                                       = 725 * 4.186 J = 3034.85 J Given translational kinetic energy = Q       ( 1/ 2) mv ^ 2 = Q from this required speed v = [ 2Q /m]                                          = 77.9 m / s (b). mass of ice m = 50g = 0.05 kg mass of water m ' = 110g = 0.11 kg mass of aluminum cup M = 62 g = 0.062 kg Initial temperature of the water and cup t = 23°C The initial temperature of the ice cube t ' = ? final temp of the system T = 0 oC heat lost by water + cup = heat gain by ice (m ' c ' dt +m' L) + M C dt = m cdt ' specific heat of water c ' = 4186J/(kg·K) Specific heat of ice c = 2090 J/(kg·K) Specific heat of aluminum C = 900 J/(kg·K) The latent heat of fusion for water L = 33.5* 10 ^4J/kg dt = 23 oC dt ' = t ' plug the values we get answer

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