Two 20.0g ice cubes at -20C are placed into 275g water at 20C. Assuming no energy in transferred to or from the surroundings, what is the final temperature of the water after all the ice melts?

Answer 1

The heat (energy) from the 20ºC water will flow to the ice at -20ºC until thermal equilibrium is reached, i.e. the final temperature. To determine this amount of heat will require the use of several constants that can be looked up. The values I find are as follows, but your values may be somewhat differernt.

∆H_fusion for water = 334 J/g

Specific heat for ice (C_ice) = 2.09 J/gº

Specific heat for liquid water (C) = 4.184 J/gº

heat lost by the 20º water = heat gained by the ice cubes

Let' first deal with the heat gained by the ice cubes:

To raise the temperature from -20º to 0º, the heat needed = q = mC∆T

m = mass of ice = 20.0 g + 20.0 g = 40 g; C = specific heat for ice = 2.09 J/g; ∆T = change in temp = 20º

q = (40.0 g)(2.09 J/gº)(20º) = 1672 J

To now melt the ice that is @0º, the heat needed is q = m∆H_f

m = mass of ice = 40.0 g; ∆Hf = heat of fusion = 334 J/g

q = (40.0 g)(334 J/g) = 13360 J

To raise temperature of the melted ice @0º to final temperature = q = mC∆T

m = mass = 40.0 g; C = specific heat of melted ice = 4.184 J/gº; ∆T = final temp. - 0º

q = (40.0 g)(4.184 J/gº)(T_f - 0º) and we will end up solving for T_f, the final temperature

Now, dealing with heat lost by the 20º water:

heat lost = q = mC∆T

m = mass of water = 275 g; C = specific heat = 4.184 J/gº; ∆T = change in temp = 20º - T_f)

q = (275 g)(4.184 J/gº0)(20º - T_f)

We now can set the heat lost by water to the total heat gained by the ice, and solve for T_f

(275 g)(4.184 J/gº0)(20º - T_f) = 1672 J + 13360 J + (40.0 g)(4.184 J/gº)(T_f - 0º)

23012 - 1151T_f = 15032 + 167T_f - 0

1318T_f = 7980

T_f = 6.05º = 6º (to 1 sig.fig.) = Final Temperature