Question

Suppose that 25.14 g of ice at -8.7°C is placed in 69.18 g of water at 98.1°C in a perfectly insulated vessel. Calculate the final temperature. (The molar heat capacity for ice is 37.5 J K-1 mol-1 and that for liquid water is 75.3 J K-1 mol-1. The molar enthalpy of fusion for ice is 6.01 kJ/mol. You must answer in Kelvin, not °C.)

Answer 1

The processes involved here are

25.14g H2O (s) -------> H2O(s) -------> H2O(l) -------> H2O(l)

at -8.7 C q1 0C q1 0C q3 TC

The heat required for all these processes = q1 +q2 +q3

This heat is supplied by the process

69.18 g of H2O(l) ------> H2O(l)

at 98.1 C Q at TC

Let us calculate the heats

q1 = heat capacity of ice x moles of water xdifference in temperature

= 37.5 x [25.14/18] x8.7 =455.66 J

q2= enthalpy of fusion x moles of water

= 6.01 x1000 J x 25.14/18

=8393.67 J

q3 = heat capacity of water x moles of eater x difference in temperature

= 75.3 J x (25.14/18) x (T-273)

= 105.17(T-273)J

Q = heat capacity of water x moles of water x difference in temperature

= 75.3J x (69.18/18)x(T-371.1)

=289.4(T-371.1)J

The heat absorbed by ice = - heat given by water

Equating the heats

455.66 J + 8393.67 J +105.17(T-273)J =-289.4(T-371.1)J

= 289.4( 371.1 -T)

Solving for T we get final temperature t = 317.45 K