Question

Question 2 (3 points)

(07.02 MC)
Given the following data for water:
Heat of fusion = 334 J/g
Heat of vaporization = 2,256J/g
Specific heat of solid = 2.09 J/g °C)
Specific heat of liquid = 4.184J/g °C)
Specific seat of gas = 1.84 J/g °C)
Calculate how much energy is needed to change 100.0 grams of liquid water at 75.0°C to vapor at 225.0°C (3 points)
A 12,500)
B 36,000)
C 259,000)
D 324,000J

Answer 1

259,000 Joules

Step-by-step explanation:

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;

Step 1: Heat energy required to raise the temp of water from 75°C to 100°C

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

Step 2: Heat required to change water at 100°C to vapor at 100°C

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

Step 3: Heat energy required to raise the temperature of ice from 100°C to 225°C.

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

Step 4: Total heat energy required

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