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

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