What are the Kilojoules needed to melt 24.0g of ice at O degrees C, warm the liquid to 100 degree C and change it to steam at 100 degree C?

Thank you,

add the heats.

melt ice: mass*heatfusionice
water : mass*spcificheatwater(100-0)
make steam: mass*heatvaporization

Well, let's break it down step by step, shall we?

To melt the ice at 0 degrees C, we need to provide the heat of fusion. The heat of fusion for ice is approximately 334 kilojoules per kilogram. So, first, let's convert 24.0 grams of ice to kilograms:

24.0 grams = 0.024 kilograms

Now, we can calculate the heat required to melt the ice:

Heat required = mass × heat of fusion
= 0.024 kg × 334 kJ/kg
= 8.016 kilojoules (approximately)

Next, we need to warm the liquid water from 0 degrees C to 100 degrees C. The specific heat capacity of water is about 4.186 kilojoules per kilogram per degree Celsius. So, using the same mass of 0.024 kilograms:

Heat required = mass × specific heat capacity × change in temperature
= 0.024 kg × 4.186 kJ/kg·°C × (100 °C - 0 °C)
= 9.9864 kilojoules (approximately)

Finally, we need to change the water to steam at 100 degrees C. The heat of vaporization for water is approximately 2260 kilojoules per kilogram. So, again using the same mass:

Heat required = mass × heat of vaporization
= 0.024 kg × 2260 kJ/kg
= 54.24 kilojoules (approximately)

Adding up all the steps:

8.016 kJ + 9.9864 kJ + 54.24 kJ = 72.2424 kilojoules (approximately)

So, it would take approximately 72.2424 kilojoules of energy to complete all these steps. But hey, who's counting?

To determine the kilojoules needed to carry out each step, we need to consider the specific heat capacity and the heat of fusion/vaporization for water. The specific heat capacity of ice is 2.09 J/g°C, the heat of fusion for ice is 334 J/g, the specific heat capacity of liquid water is 4.18 J/g°C, and the heat of vaporization for water is 2260 J/g.

Step 1: Melting ice at 0°C
The energy required to melt the ice can be calculated using the formula:
Energy = mass × heat of fusion

First, convert the mass of ice from grams to kilograms:
Mass of ice = 24.0g = 0.024 kg

Calculate the energy required:
Energy = 0.024 kg × 334 J/g = 7.98 kJ

So, 7.98 kJ of energy is needed to melt the ice.

Step 2: Warming the liquid water from 0°C to 100°C
The energy required to warm the liquid water can be calculated using the formula:
Energy = mass × specific heat capacity × temperature change

The mass of water remains the same at 24.0g (or 0.024 kg).

Calculate the energy required:
Energy = 0.024 kg × 4.18 J/g°C × (100°C - 0°C) = 9.97 kJ

So, 9.97 kJ of energy is needed to warm the liquid water.

Step 3: Changing liquid water to steam at 100°C
The energy required to change liquid water to steam can be calculated using the formula:
Energy = mass × heat of vaporization

The mass of liquid water remains the same at 24.0g (or 0.024 kg).

Calculate the energy required:
Energy = 0.024 kg × 2260 J/g = 54.24 kJ

So, 54.24 kJ of energy is needed to change the liquid water to steam.

Summing up the energies needed for each step, the total energy required is:
7.98 kJ (melting ice) + 9.97 kJ (warming liquid water) + 54.24 kJ (changing liquid water to steam) = 72.19 kJ

Therefore, the kilojoules needed to melt 24.0g of ice at 0°C, warm the liquid to 100°C, and change it to steam at 100°C is approximately 72.19 kJ.

To calculate the total kilojoules needed to melt the ice, warm the liquid, and change it to steam, we need to consider each step separately and then add them up.

First, we need to calculate the energy required to melt the ice. The specific heat of fusion for ice is 334 J/g. Since we have 24.0 grams of ice, we can calculate the energy required for melting the ice as follows:

Energy = Mass x Specific Heat of Fusion
Energy = 24.0 g x 334 J/g

Next, we need to calculate the energy required to warm the liquid from 0 degrees Celsius to 100 degrees Celsius. The specific heat capacity of water is 4.184 J/g°C. The energy required can be calculated as:

Energy = Mass x Specific Heat Capacity x Temperature Change
Energy = 24.0 g x 4.184 J/g°C x (100°C - 0°C)

Finally, we need to calculate the energy required to change the liquid water to steam. The heat of vaporization for water is 2260 J/g. The energy can be calculated as:

Energy = Mass x Heat of Vaporization
Energy = 24.0 g x 2260 J/g

Now, let's add up all the energies to get the total energy required:

Total Energy = Energy to melt ice + Energy to warm liquid + Energy to change to steam

You can substitute the calculated values we discussed earlier into this equation to get the final answer.