Find the amount of heat required to convert 10g of ice to steam at 110 degree Celsius.

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To find the amount of heat required to convert 10 grams of ice to steam at 110 degrees Celsius, we need to consider the different phase changes involved.

Here are the steps to calculate the amount of heat:

1. Calculate the heat required to raise the temperature of ice from -10 degrees Celsius to 0 degrees Celsius.

The specific heat capacity of ice is approximately 2.09 J/g°C. Therefore, the heat required is:

Q1 = mass × specific heat capacity × temperature change
= 10g × 2.09 J/g°C × (0 - (-10) °C)
= 10g × 2.09 J/g°C × 10 °C
= 209 J

2. Calculate the heat required to convert the ice at 0 degrees Celsius to water at 0 degrees Celsius.

The heat of fusion (enthalpy of fusion) for ice is approximately 333.5 J/g. Therefore, the heat required is:

Q2 = mass × heat of fusion
= 10g × 333.5 J/g
= 3335 J

3. Calculate the heat required to raise the temperature of water from 0 degrees Celsius to 100 degrees Celsius.

The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, the heat required is:

Q3 = mass × specific heat capacity × temperature change
= 10g × 4.18 J/g°C × (100 - 0) °C
= 10g × 4.18 J/g°C × 100 °C
= 4180 J

4. Calculate the heat required to convert the water at 100 degrees Celsius to steam at 100 degrees Celsius.

The heat of vaporization (enthalpy of vaporization) for water is approximately 2260 J/g. Therefore, the heat required is:

Q4 = mass × heat of vaporization
= 10g × 2260 J/g
= 22600 J

5. Calculate the heat required to raise the temperature of steam from 100 degrees Celsius to 110 degrees Celsius.

The specific heat capacity of steam is approximately 2.03 J/g°C. Therefore, the heat required is:

Q5 = mass × specific heat capacity × temperature change
= 10g × 2.03 J/g°C × (110 - 100) °C
= 10g × 2.03 J/g°C × 10 °C
= 203 J

Now, add up all the calculated values to find the total heat required:

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 209 J + 3335 J + 4180 J + 22600 J + 203 J
≈ 30427 J

Therefore, approximately 30427 Joules of heat are required to convert 10 grams of ice to steam at 110 degrees Celsius.

To find the amount of heat required to convert 10g of ice to steam at 110 degrees Celsius, we need to consider the different stages of the phase change process.

First, we need to calculate the heat required to raise the temperature of the ice from its initial temperature (let's assume it's at 0 degrees Celsius) to its melting point (0 degrees Celsius). The heat required for this step can be calculated using the formula:

Q1 = mass × specific heat capacity × temperature change

The specific heat capacity of ice is approximately 2.09 J/g°C.

So, for the ice to reach the melting point, the calculation would be:

Q1 = 10g × 2.09 J/g°C × (0°C - 0°C) = 0 J (no heat is needed, as the ice is already at its melting point)

Next, we need to calculate the heat required to melt the ice at its melting point. The heat required for this phase change can be calculated using the formula:

Q2 = mass × heat of fusion

The heat of fusion for ice is approximately 334 J/g.

So, for the ice to melt completely, the calculation would be:

Q2 = 10g × 334 J/g = 3340 J

After the ice has melted, we need to calculate the heat required to raise the temperature of the water from its melting point (0 degrees Celsius) to the desired final temperature (110 degrees Celsius). The specific heat capacity of water is approximately 4.18 J/g°C (assuming it is still in the liquid state).

So, for the water to reach the final temperature, the calculation would be:

Q3 = 10g × 4.18 J/g°C × (110°C - 0°C) = 4,598 J

Lastly, we need to calculate the heat required to convert the water at its boiling point (100 degrees Celsius) to steam at the desired final temperature (110 degrees Celsius). The heat required for this phase change can be calculated using the formula:

Q4 = mass × heat of vaporization

The heat of vaporization for water is approximately 2260 J/g.

So, for the water to completely evaporate into steam, the calculation would be:

Q4 = 10g × 2260 J/g = 22,600 J

To find the total heat required, we simply add up all of the individual heat values:

Total heat = Q1 + Q2 + Q3 + Q4
Total heat = 0 J + 3340 J + 4,598 J + 22,600 J
Total heat = 29,538 J

Therefore, the amount of heat required to convert 10g of ice to steam at 110 degrees Celsius is approximately 29,538 J.

d question is not complete.either d (specific heat capacity of ice) is give or(specific heat of fusion of ice)is given so check d question.