An amount of ice at 0°C is added to an 8.25 kg tub of water at 25.0°C. If their final temperature is 18.2°C, what mass of ice was added?

Answer 1

How much heat was lost to change the temp of water from 25º to 18.2º?

q = mC∆T = (8250 g)(4.184 J/gº)(6.8º) = 234,722 J = -234.7 kJ of heat lost by the water

This amount of heat had to go to the ice.

First, we have to melt the ice at 0ºC:

Then to change the temperature of the liquid ice at 0º to 18.2ºC

To melt the ice: q = m∆Hfusion = 334 kJ/kg m

To change temp from 0º to 18.2ºC: q = mC∆T = m(4.184 kJ/kgº)(18.2º)

total heat = 334 kJ/kg (m) + 76m

234.7 kJ = 334m + 76m = 410m

m = 0.572 kg of ice

m = 572 g of ice

(NOE: be sure to check my math for possible errors)