Question

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK and liquid water = 4.184 J/gK.1.31 × 105 J2.14 × 104 J1.66 × 104 J3.50 × 104 J6.59 × 103 J

Answer 1

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.