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

3,50x10⁴J

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. 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!

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.