PROBLEM 103P

An ice cube at 0.00 °C with a mass of 23.5 g is placed into 550.0 g of water, initially at 28.0 °C, in an insulated container. Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all of the ice has melted?

Solution 103P

Given:

Mass of ice = 23.5g

Mass of water = 550 g

Initial temperature = 28oC

We will have to find the temperature of the entire water sample after all of the ice has melted?

Explanation:

The specific enthalpy of fusion of water is 333.55 kJ/kg at 0 °C. = 333.55 J g-1deg-1

Heat gained by the ice/water = heat lost by warm water Q = mcΔT

Q is the quantity of heat,

m = mass of water in g,

c is specific heat in J g-1 deg-1 (= 4.184 for H2O) and ΔT is the temp change. Let final T be T

Heat gained by ice = m(Hfus) + mcΔT

Heat gained by ice = 333.55×23.5 + 23.5×4.184×(T-0.0) = 7838 +98.324T

heat lost by warm water = 550×4.184×(28.0 - T) = 64433 – 2301T

Hence 7838 +98.324T = 64433 – 2301T

98.324T + 2301T = 64433-7838

(98.324 + 2301)T =57265

T(3299.3) = 57265

T = 57265/3299.3 = 17.3 °C

Thus the temperature of the entire water sample after all of the ice has melted is 17.3 °C