Question

I am having trouble with part B and could use some help

Three 110.0-g ice cubes initially at 0°C are added to 0.900 kg of water initially at 21.0°C in an insulated container. (a) What is the equilibrium temperature of the system? °C 0 (b) What is the mass of unmelted ice, if any, when the system is at equilibrium? 93.75 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kg

Answer 1

I think you have done the problem correctly only you have not converted the answer into kg.

The ice is converted into water at 0°C decreasing the temperature of the 0.900 kg of water which is initially at 21°C. The amount of heat released by the water when its temperature changes from 21°C to 0°C is

$Q=mC(21^oC-0^oC)$

$Q=(0.900\ kg)\times (4186\ J/(kg ^oC))\times(21^oC)$

The mass of ice that heat Q can melt is

$M=\frac{Q}{L_{fusion}}$

$M=\frac{79115\ J}{334000\ J/kg}$

There were initially three ice cubes each of mass 110g. The initial mass of the ice is

$M_o=3\times 110\ g=330\ g=0.330\ kg$

The mass of unmelted ice is

$\Delta M=M_o-M$

$\Delta M=0.330\ kg-0.2369\ kg$

$\boxed {\Delta M=0.0931\ kg}$