Question

A mixture containing 21.4g of ice (at 0.0 degrees C) and 75.3 g of water (55.3 degrees C) is placed in an insulated container. Assuming no loss of heat to the surroundings, what is the final temp of the mixture ?

Answer 1

Final temp of the mixture wil be 25.4 ⁰C

The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder (q_lost = q_gain), we have this

heat to melt ice + heat to warm cold water by unknown amount = heat lost by the warm water

[(6020)(21.4/18.0)] + [(21.4)(x - 0)(4.184)] = (75.30)(55.3 - x)(4.184)

I used 6020 J rather than 6.02 kJ for the molar heat of fusion. That is because the other two parts of the left-hand side of the equation will give Joules as their answer. I used 6020 so that all three parts would be in Joules. If I had used 6.02, then that middle part woud have been in units of kJ.

specific heat of water = 4.18 J/(g * ˚C)

[(6020)(1.188)] + 89.536X = 315.0552   (55.3 - x)

7157 + 89.536X + 315.0552 X = 17422

404.5912 X = 17422 - 7157

404.5912 X = 10264

                X = 25.37 ⁰C