How many joules of heat are needed to change 50.0 grams of ice at -15.0 °C to steam at 120.0 °C?

1 Answer
Dec 14, 2014

The answer is 153.7kJ153.7kJ.

What you are asked to determine is the total energy required to go from ice to water, and then from water to vapor - the phase changes underwent by the water molecules.

In order to do this, you'll need to know:

Heat of fusion of water: DeltaH_f = 334 J/g;
Heat of fusion vaporization of water: DeltaH_v = 2257 J/g;
Specific heat of ice: c = 2.09 J/g^@C;
Specific heat of water: c = 4.18 J/g^@C;
Specific heat of steam: c = 2.09 J/g^@C;

So, the following steps describe the overall process:

1. Determine the heat required to raise the temperature of the ice from -15.0^@C to 0^@C:

q_1 = m* c_(ice) * DeltaT = 50.0g * 2.09 J/(g*^@C) * (0^@C - (-15^@C)) = 1567.5 J

2. Determine the heat required to convert 0^@C ice to 0^@C water:

q_2 = m * DeltaH_(f) = 50.0 g * 334 J/(g) = 16700J

3. Determine the heat required to go from water at 0^@C to water at 100^@C:

q_3 = m * c_(water) * DeltaT = 50.0g * 4.18 J/(g*^@C) * (100^@C - 0^@C) = 20900J

4. Determine the heat required to convert 100^@C water to 100^@C vapor:

q_4 = m * DeltaH_(v) = 50.0g * 2257 J/(g) = 112850 J

5. Determine the heat required to go from 100^@C vapor to 120^@C vapor:

q_5 = m * c_(vap o r) * DeltaT = 50.0g * 2.09 J/(g*^@C) * (120^@C - 100^@C) = 2090J

Therefore, the total heat required is

q_(TOTAL) = q_1+q_2+q_3+q_4+q_5 = 152696.5J = 153.7kJ