How much heat (in Kj ) is required to warm 10.0g  of ice, initially at -11.0 C, to steam at 112.0 C?

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To calculate the amount of heat required to warm the ice to steam, we need to consider the following steps:

1. Calculate the heat required to warm the ice from -11.0°C to 0°C.
2. Calculate the heat required to melt the ice at 0°C.
3. Calculate the heat required to warm the water from 0°C to 100°C.
4. Calculate the heat required to vaporize the water at 100°C.
5. Calculate the heat required to warm the steam from 100°C to 112.0°C.

We will use the specific heat capacity (C) and the heat of fusion (Hf) and vaporization (Hv) for water in these calculations.

Step 1: Calculate heat required to warm ice from -11.0°C to 0°C.
The specific heat capacity of ice is 2.09 J/g°C.
q1 = mass × specific heat capacity × temperature change
= 10.0 g × 2.09 J/g°C × (0°C - (-11.0°C))
= 10.0 g × 2.09 J/g°C × 11.0°C
= 230.9 J

To convert joules to kilojoules, we divide by 1000.
q1 = 230.9 J / 1000
≈ 0.231 kJ

Step 2: Calculate heat required to melt ice at 0°C.
The heat of fusion of ice is 334 J/g.
q2 = mass × heat of fusion
= 10.0 g × 334 J/g
= 3340 J

To convert joules to kilojoules, we divide by 1000.
q2 = 3340 J / 1000
= 3.34 kJ

Step 3: Calculate heat required to warm water from 0°C to 100°C.
The specific heat capacity of water is 4.18 J/g°C.
q3 = mass × specific heat capacity × temperature change
= 10.0 g × 4.18 J/g°C × (100°C - 0°C)
= 4180.0 J

To convert joules to kilojoules, we divide by 1000.
q3 = 4180.0 J / 1000
= 4.18 kJ

Step 4: Calculate heat required to vaporize the water at 100°C.
The heat of vaporization of water is 2260 J/g.
q4 = mass × heat of vaporization
= 10.0 g × 2260 J/g
= 22600 J

To convert joules to kilojoules, we divide by 1000.
q4 = 22600 J / 1000
= 22.6 kJ

Step 5: Calculate heat required to warm the steam from 100°C to 112.0°C.
The specific heat capacity of steam is 2.01 J/g°C.
q5 = mass × specific heat capacity × temperature change
= 10.0 g × 2.01 J/g°C × (112.0°C - 100°C)
= 240.0 J

To convert joules to kilojoules, we divide by 1000.
q5 = 240.0 J / 1000
= 0.24 kJ

Now, to get the total heat, we sum up all the individual heats:
Total heat = q1 + q2 + q3 + q4 + q5
= 0.231 kJ + 3.34 kJ + 4.18 kJ + 22.6 kJ + 0.24 kJ
≈ 30.64 kJ

Therefore, approximately 30.64 kJ of heat is required to warm 10.0g of ice initially at -11.0°C to steam at 112.0°C.