Question

How much heat would need to be removed to cool 212.3 g of water from 25.6°C to -10.7°C?

Answer 1

from 25.6°C to 0°C liquid state water Calculating energy: mass of water = m = 212.3 g Specific heat of water = S = 4.184 J/gC Temperature difference = AT = 25.60 - 0.00 = 25.60 °C Heat released by water = m S AT Heat released by water = 212.3 * 4.184 x 25.60 Heat released by water = 22739.5 J | Calculating number of moles: mass of water = 212.3 g molar mass of water = 18.00 g/mol mole of water = ? mol mass Mole of water = molar mass 212.3 g Mole of water = 18.00 g/mol Mole of water = 11.794 mol at 0°C from liquid state water to solid state ice

Calculating energy: moles of water = n = 11.794 g heat of fusion water = AHfus = 6.010 kJ/mol Heat released by water = n AHfus Heat released by water = 11.794 x 6.010 Heat released by water = 70.8846 kJ Heat released by water = 70884.6 J from 0°C to -10.7°C solid state ice Calculating energy: mass of ice Specific heat of ice Temperature difference Heat released by ice Heat released by ice Heat released by ice = m = 212.3 g = Cp = 2.090 J/gC = AT = 0.00 - -10.70 = 10.70 °C = m S AT = 212.3 x 2.090 * 10.70 = 4747.7 J Calculating total energy: 70884.6 + 4747.7 total energy released total energy released total energy released = - = 22739.5 + 98371.8 J 98.4 kJ