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 J
2.14 × 104 J
1.66 × 104 J
3.50 × 104 J
6.59 × 103 J

Answer 1

3,50x10⁴J

Step-by-step explanation:

To obtain the energy required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C you need to caclulate:

1. Energy to increase temperature of ice from -30,0°C to 0,0°C
2. Heat of fusion (Change of solid to liquid water)
3. Energy yo increase the temperature of liquid water from 0,0°C to 73,0°C

1. The energy is obtained with specific heat capacity of ice, thus:

Q = C×m×ΔT

Q = 2,06 J/gK×50,0g×(0,0°C- -30,0)

Q = 3090 J

2. The heat of fusion is:

333 J/g×50g = 16650J

3. Energy to increase temperature of liquid water is:

Q = C×m×ΔT

Q = 4,184 J/gK×50,0g×(73,0°C - 0,0°C)

Q = 15272 J

Thus, total heat is:

15272J + 16650J + 3090J = 35012J = 3,50x10⁴J

I hope it helps!