Question

A 25 g piece of hot iron is dropped into a glass of ice water containing 15.0 g of ice and 50.0 g of water. If the iron loses 8.00 kJ of energy to the ice water in achieving equilibrium, what is the equilibrium temperature? ∆Hfus = 6.01 kJ/mol

Answer 1

The amount of heat required to melt ice , q = m∆H_fus

Where

m = mass of ice = 15.0 g

∆H_fus = 6.01 kJ/mol

          = 6.01 kJ/mol x (mol / 18.0 g)

          = 0.334 kJ/g

So q = 15.0g x 0.334 kJ

        = 5.01 kJ

So among 8.00 kJ of heat lost by iron , 5.01 kJ of heat is utilized to melt ice to water.

The remaining 8.00-5.01 = 2.99 kJ = 2.99x1000 J = 2990 J of heat is utilized to convert water( of mass 15.0g) & water ( of mass 50.0g) to a common temperature t .

So Q = 2990J = mcdt + m'cdt

                       = (m+m')cdt

Where

m = mass of ice = 15.0 g

m' = mass of water = 50.0 g

c = specific heat capacity of water = 4.186 J/g degree C

dt = change in temperature = final - initial

    = t - 0

    = t oC

Plug the values we get

2990 = ( 15.0+50.0) x 4.186xt

t = 10.99 ^oC

Therefore the equilibrium temperature is 10.99 ^oC