Answer:
There is 3.5*10^4 J of energy needed.
Step-by-step 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.