Answer to Question #56165 in General Chemistry for tracy guido

Question

Calculate the kJ of heat energy required to convert 155.0 grams of ice at -44.0 C to all steam at 100.0 C. Assume that no energy in the form of heat is transferred to the environment. (Constants, only some of which are needed: Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: liquid water = 4.184 J/g K, steam = 1.92 J/g K, ice = 2.06 J/g K)

Answer 1

The energy needed to provide this change consists of several parts:
1) Energy used to heat an ice from -44 oC to 0 oC:
Q1 = C1m∆T, where C1 – the specific heat capacity of ice, m – the mass and ∆T – the change of temperature.
Q1 = 2.06 J/(g K) × 155 g × 44 K = 14049.2 J
2) Energy used for fusion of an ice:
Q2 = C2m, where C1 – the heat of fusion of ice, m – the mass.
Q2 = 333 J/g × 155 g = 51615 J
3) Energy used to heat water from 0 oC to 100 oC:
Q3 = C3m∆T, where C1 – the specific heat capacity of water, m – the mass and ∆T – the change of temperature.
Q3 = 4.184 J/(g K) × 155 g × 100 K = 64852 J
4) Energy used for evaporation of water:
Q4 = C4m, where C4 – the heat of vaporization, m – the mass.
Q4 = 2256 J/g × 155 g = 349680 J
Finally, the total energy equals:
Q = Q1+ Q2+ Q3+ Q4= 14049.2 J + 51615 J + 64852 J + 349680 J = 480.196 kJ