What mass of water at 15c can be cooled 1c by heat necessary to melt 185g of ice at 0c?

Answer 1

The heat required to melt 185 g of ice at 0 C is given by :

QICE = ( mICE ) ( hsfICE ) = ( 185 g ) ( 333.7 J / g ) = 61734.5 J

QW = - QICE = ( mW ) (CshW ) ( - 1.0 C deg )

mW = ( - 61734.5 J ) / ( 4.184 J / g - C deg ) ( - 1.0 C deg )

mW = 14,750 g = 14.75 kg of water <-----

where

mICE is mass of ice

hsfICE is (specific) enthalpy of solid fusion, i.e. enthalpy per unit mass

Cshw is the "heat capacity" of water,

mW is mass of water

NOTE: the literature value for water at °C is actually 4.1858 J/g/°C, so the actual answer should be closer to 14.7485 (not exactly though since the temperature change would be from 15 °C to 14 °C so you would need the heat capacity as a function of temperature over that range, not just the isothermal value at 15 °C). Within the justifiable precision, it still rounds to 14.75 kg, just like the answer above so the error is negligible.

Answer 2

To cool 1 g of water by 1°C, 4.18 J of heat is required. 185 g of ice requires 185 g × 334 J/g = 61790 J of heat to melt. To cool the water from 15°C to 14°C, it will need (15-14) * mass of water * 4.18 J. Solving these equations will give you the mass of water needed.