Calorimetry Question. How to solve it?

A cube of iron (density = 8000 kg/m^3, specific heat capacity = 470 J/kg-K) is heated to a high temperature and is placed on a large block of ice at 0^oC. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temp of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg/m^3 and the latent heat of fusion of ice = 3.36 * 10^5 J/Kg.

2 Answers
A08
Feb 10, 2018

Second approach where it is assumed that water after melting is displaced and stays inside the cavity formed in which iron cube also sinks in. Density of water taken as #1000\ kgm^-3#.
.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Let #v\ m^3# be the volume of iron cube.
Volume of #1\ kg# of ice#=1/900\ m^3#
Volume of #1\ kg# of water thus formed after melting of ice#=1/1000\ m^3#
Difference in volume#=(1/900-1/1000)\ m^3#
Amount of ice required to melt to accommodate #v\ m^3# of volume of iron cube#=v/(1/900-1/1000)\ kg#
#=(vxx9000)\ kg#

Heat lost by the iron cube#=msDeltaT=(vxx8000)xx(470)xxDeltaT#

Heat gained by ice for melting from ice at #0^@\ C# to water at #0^@\ C# #=mL=(vxx9000)xx(3.36 xx 10^5)#

Using Law of conservation of energy, equating the two we get
#(vxx8000)xx(470)xxDeltaT=(vxx9000)xx(3.36 xx 10^5)#
#=>DeltaT=(vxx9000)xx(3.36 xx 10^5)xx1/((vxx8000)xx(470))=804#
Since final equilibrium temperature #=0^@\ C#
Therefore, initial temperature of iron cube#=804^@\ C#

A08
Feb 10, 2018

First approach where it is assumed that water after melting is displaced and expelled out of the cavity formed in which iron cube sinks in.
.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Let #v\ m^3# be the volume of iron cube.

Amount of ice required to melt to accommodate #v\ m^3# of volume of iron cube#=(vxx900)\ kg#

Heat lost by the iron cube#=msDeltaT=(vxx8000)xx(470)xxDeltaT#

Heat gained by ice for melting from ice at #0^@\ C# to water at #0^@\ C# #=mL=(vxx900)xx(3.36 xx 10^5)#

Using Law of conservation of energy, equating the two we get
#(vxx8000)xx(470)xxDeltaT=(vxx900)xx(3.36 xx 10^5)#
#=>DeltaT=(vxx900)xx(3.36 xx 10^5)xx1/((vxx8000)xx(470))=80.4#
Since final equilibrium temperature #=0^@\ C#
Therefore, initial temperature of iron cube#=80.4^@\ C#

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