Using the values for the heat of fusion, specific heat of water, or heat of vaporization, calculate the amount of heat energy in each of the following:

* calories removed to condense 125g of steam at 100 degree C to cool the liquid to 15 degree C

* joules needed to melt a 525-g ice cube at 0 degree C and to war the liquid to 15 degree C

*Kilocalories removed to condense 85 g of steam at 100 degree C, cool the liquid to 0 degree C, and freeze it at 0 degree C

First one.

q1 = heat removed to condense 125 g steam at 100 C to water at 100 C.
q1 = mass steam x heat vaporization.

q2 = heat removed to cool liquid water at 100 C to 15 C.
q2 = mass water x specific heat water x (Tfinal-Tinitial).

qtotal = q1 + q2

300

500kl

To calculate the amount of heat energy in each of the given scenarios, we need to use the equations involving the heat of fusion, specific heat, and heat of vaporization.

1. Calories removed to condense 125g of steam at 100°C to cool the liquid to 15°C:

To condense steam at 100°C, we need to use the equation:

Heat energy = mass * heat of vaporization

Given:
Mass of steam = 125g
Heat of vaporization of water = 540 cal/g

Heat energy to condense the steam = 125g * 540 cal/g = 67,500 calories

To cool the liquid to 15°C, we need to use the equation:

Heat energy = mass * specific heat * change in temperature

Given:
Mass of liquid water = 125g
Specific heat of water = 1 cal/g°C
Change in temperature = (100 - 15)°C = 85°C

Heat energy to cool the liquid = 125g * 1 cal/g°C * 85°C = 10,625 calories

Total amount of heat energy = Heat energy to condense steam + Heat energy to cool liquid
= 67,500 calories + 10,625 calories
= 78,125 calories

Therefore, the amount of heat energy required is 78,125 calories.

2. Joules needed to melt a 525g ice cube at 0°C and warm the liquid to 15°C:

To melt the ice cube, we need to use the equation:

Heat energy = mass * heat of fusion

Given:
Mass of ice = 525g
Heat of fusion of water = 334 J/g

Heat energy to melt the ice = 525g * 334 J/g = 174,900 joules

To warm the liquid to 15°C, we need to use the equation:

Heat energy = mass * specific heat * change in temperature

Given:
Mass of liquid water = 525g
Specific heat of water = 4.18 J/g°C
Change in temperature = (15 - 0)°C = 15°C

Heat energy to warm the liquid = 525g * 4.18 J/g°C * 15°C = 34,798.5 joules

Total amount of heat energy = Heat energy to melt ice + Heat energy to warm liquid
= 174,900 joules + 34,798.5 joules
= 209,698.5 joules

Therefore, the amount of heat energy required is 209,698.5 joules.

3. Kilocalories removed to condense 85g of steam at 100°C, cool the liquid to 0°C, and freeze it at 0°C:

To condense steam at 100°C, we need to use the equation:

Heat energy = mass * heat of vaporization

Given:
Mass of steam = 85g
Heat of vaporization of water = 540 cal/g

Heat energy to condense the steam = 85g * 540 cal/g = 45,900 calories

To cool the liquid to 0°C, we need to use the equation:

Heat energy = mass * specific heat * change in temperature

Given:
Mass of liquid water = 85g
Specific heat of water = 1 cal/g°C
Change in temperature = (100 - 0)°C = 100°C

Heat energy to cool the liquid = 85g * 1 cal/g°C * 100°C = 8,500 calories

To freeze the liquid at 0°C, we need to use the equation:

Heat energy = mass * heat of fusion

Given:
Mass of liquid water = 85g
Heat of fusion of water = 334 J/g

Heat energy to freeze the liquid = 85g * 334 J/g * (1 cal/4.184 J) = 679 calories

Total amount of heat energy = Heat energy to condense steam + Heat energy to cool liquid + Heat energy to freeze liquid
= 45,900 calories + 8,500 calories + 679 calories
= 55,079 calories

Therefore, the amount of heat energy required is 55,079 calories.