How much heat is required in order to change 5 g of ice at -15 degree to steam at 150 degree?

To calculate the amount of heat required to change a substance from one state to another, you need to consider three separate processes:

1. Heating the ice from -15°C to 0°C.
2. Melting the ice to water at 0°C.
3. Heating the water from 0°C to 100°C (boiling point).
4. Vaporizing the water to steam at 100°C.
5. Heating the steam from 100°C to 150°C.

Let's calculate the amount of heat required for each step:

1. Heating the ice from -15°C to 0°C:
The specific heat capacity of ice is 2.09 J/g°C.
The temperature change is 0 - (-15) = 15°C.
The mass of the ice is 5 g.

The heat required = mass × specific heat capacity × temperature change
= 5 g × 2.09 J/g°C × 15°C
= 156.75 J

2. Melting the ice to water at 0°C:
The heat of fusion for ice is 334 J/g.
The mass of the ice is still 5 g.

The heat required = mass × heat of fusion
= 5 g × 334 J/g
= 1670 J

3. Heating the water from 0°C to 100°C:
The specific heat capacity of water is 4.18 J/g°C.
The temperature change is 100°C - 0°C = 100°C.
The mass of the water is now 5 g.

The heat required = mass × specific heat capacity × temperature change
= 5 g × 4.18 J/g°C × 100°C
= 2090 J

4. Vaporizing the water to steam at 100°C:
The heat of vaporization for water is 2260 J/g.
The mass of the water is 5 g.

The heat required = mass × heat of vaporization
= 5 g × 2260 J/g
= 11300 J

5. Heating the steam from 100°C to 150°C:
The specific heat capacity of steam is 2.03 J/g°C.
The temperature change is 150°C - 100°C = 50°C.
The mass of the steam is still 5 g.

The heat required = mass × specific heat capacity × temperature change
= 5 g × 2.03 J/g°C × 50°C
= 509.5 J

Now, let's add up all the heat required for each step:

Total heat required = 156.75 J + 1670 J + 2090 J + 11300 J + 509.5 J
= 15726.25 J

Therefore, it would require 15726.25 Joules of heat to change 5 g of ice at -15°C to steam at 150°C.

To calculate the amount of heat required to change a substance from one state to another (like from ice to steam), you need to consider the specific heat capacities and heat of phase changes of the substance.

In this case, we need to consider three steps: heating the ice to its melting point, changing the ice to water at 0°C, heating the water from 0°C to its boiling point, and changing the water to steam.

Step 1: Heating the ice from -15°C to its melting point at 0°C
The specific heat capacity of ice is 2.09 J/g°C. Thus, to raise the temperature of 5 g of ice from -15°C to 0°C, you would use the formula:
Heat = mass × specific heat capacity × temperature change
Heat = 5 g × 2.09 J/g°C × (0°C - (-15°C))
Heat = 5 g × 2.09 J/g°C × 15°C
Heat = 156.75 J

Step 2: Melting the ice to water at 0°C
The heat of fusion for water is 333.55 J/g. This represents the amount of heat required to change one gram of ice at 0°C to one gram of water at 0°C. Since we have 5 g of ice, the heat required would be:
Heat = mass × heat of fusion
Heat = 5 g × 333.55 J/g
Heat = 1667.75 J

Step 3: Heating the water from 0°C to its boiling point at 100°C
The specific heat capacity of water is 4.18 J/g°C. Thus, to raise the temperature of 5 g of water from 0°C to 100°C, you would use the formula:
Heat = mass × specific heat capacity × temperature change
Heat = 5 g × 4.18 J/g°C × (100°C - 0°C)
Heat = 5 g × 4.18 J/g°C × 100°C
Heat = 2090 J

Step 4: Changing the water to steam at 100°C
The heat of vaporization for water is 2260 J/g. This represents the amount of heat required to change one gram of water at 100°C to one gram of steam at 100°C. Since we have 5 g of water, the heat required would be:
Heat = mass × heat of vaporization
Heat = 5 g × 2260 J/g
Heat = 11300 J

Now, add up the heat from all the steps to get the total heat required:
Total Heat = 156.75 J + 1667.75 J + 2090 J + 11300 J
Total Heat = 15214.5 J

Therefore, approximately 15214.5 J of heat is required to change 5 g of ice at -15°C to steam at 150°C.

Look up the specific heats of ice, water and steam. Call them Ci, Cw and Cs.

Cw is 1.00 cal/g C

You will also need the heat of fusion
Hf = 80 cal/g
and the heat of vaporization
Hv = 540 cal/g

Q required = 5g * [15Ci + 100Cw + 50 Cs + Hf + Hv] calories