Note: Please show all work and calculation setups to get full credit. T. he following may be used on this assignment: specific heat of (water=4.184 J/g oC; ice=2.03 J/g oC; steam=1.99 184 J/g oC); heat of fusion of water=80. cal/g; heat of vaporization=540 cal/g; 1cal=4.184J. Calculate the energy required (in J) to convert 25 g of ice at -15 oC to water at 75 oC.

16974J of energy are required

Explanation:

The energy required is:

* The energy to heat solid water from -15°C to 0°C using:

q = m*S*ΔT

* The energy to convert the solid water to liquid water:

q = dH*m

* The energy required to increase the temperature of liquid water from 0°C to 75°C

q = m*S*ΔT

The first energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of ice = 2.03J/g°C

ΔT is change in temperature = 0°C - (-15°C) = 15°C

q = 25g*2.03J/g°C*15°C

q = 761.3J

The second energy is:

q = dH*m

m = Mass water = 25g

dH is heat of fusion of water = 80cal/g

q = 80cal/g*25g

q = 2000cal * (4.184J/1cal) = 8368J

The third energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of water= 4.184J/g°C

ΔT is change in temperature = 75°C-0°C = 75°C

q = 25g*4.184J/g°C*75°C

q = 7845J

The energy is: 7845J + 8368J + 761J =

16974J of energy are required

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.