A beaker contains 0.500 kg of hot water at T = 60 degree C. Ice of mass 0.250 kg

Question

A beaker contains 0.500 kg of hot water at T = 60 degree C. Ice of mass 0.250 kg and temperature T = - 20 degree C is then added to the beaker. Assume that the system is completely insulated from its surroundings. What will the final phase of the mixture be? Explicitly show how you arrive at your conclusion. You are now given another container of 0.500 kg of hot water at T = 60 degree C. Ice of mass 0.250 kg at temperature T = 0 degree C is then added to the beaker. Calculate the final equilibrium temperature of the system.

Explanation / Answer

specific heat of water = 4.179

ice = 2.03

energy required by ice to get converted to water

= mL + msdT

= 0.25*1000* (334 ) + 0.25*1000* 2.03 *(20)

=93650 J

heat contained in water

=msdT => 0.500*1000*4.179 *60 =125.37 *1000 =125370 J

since water is having more heat than needed by ice

so whole ice will get converted to water

final phase will be water

b) heat in water = msdT = 125370 J

latent heat of fusion = mL =0.25*1000* (334 )  =83500 J

so remaining heat after whole ice gets converted t water = 125370 - 83500

=41870 J

this remaining heat will increase total water's temp

let that be T

41870 = (0.5+0.25)*1000 * 4.179 * dT

T = 13.358857781 deg C

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