<h3>
Answer:</h3>
259,000 Joules
<h3>
Explanation:</h3>
To calculate the heat energy needed to change 100 grams of liquid water at 75°C to vapor at 225°C, we will do this in steps;
<h3>
Step 1: Heat energy required to raise the temp of water from 75°C to 100°C</h3>
Mass of water = 100 g
Heat capacity of water = 4.184J/g °C
Temperature change (75°C to 100°C) = 25°C
Using the formula to get Quantity of heat;
Q = mcΔT
= 100 g × 4.184 J/g °C × 25°C
= 10,460 joules
<h3>
Step 2: Heat required to change water at 100°C to vapor at 100°C</h3>
Mass of water = 100 g
Heat of vaporization = 2,256 J/g
Heat required to change water to vapor without a change in temperature is given by;
Q = m × Lv
= 100 × 2,256 J/g
= 225,600 Joules
<h3>
Step 3: Heat energy required to raise the temperature of ice from 100°C to 225°C.</h3>
Mass of vapor = 100 g
Specific heat of gas = 1.84 J/g °C
Change in temperature (100°C to 225°C) = 125°C
Therefore;
Q = mcΔT
= 100 g × 1.84 J/g °C × 125°C
= 23,000 joules
<h3>Step 4: Total heat energy required</h3>
Total energy = 10,460 J + 225,600 J + 23,000 J
= 259,060 J
= 259,000 J(approx)
Therefore, heat energy required is 259,000 Joules
