how much heat kj is requried to warm 10.0 grams of ice , initiall at -10.0c to steam at 110.0 c

Answers

Answer 1

The amount of heat required to warm 10.0 grams of ice from -10.0°C to steam at 110.0°C is 29,513 J or 29.5 kJ.

To solve this problem, we need to break it down into several steps, since the heat required to warm the substance depends on its phase and temperature.

Heating the ice from -10.0°C to 0°C

The first step is to heat the ice from its initial temperature of -10.0°C to its melting point at 0°C. To do this, we need to calculate the heat required using the formula;

Q = m × C × ΔT

where Q is heat required, m is mass of the substance, C is specific heat capacity of the substance, and ΔT is the change in temperature.

The specific heat capacity of ice will be 2.09 J/g°C, so;

Q₁ = 10.0 g × 2.09 J/g°C × (0°C - (-10.0°C)) = 209 J

Melting the ice at 0°C

Next, we need to calculate the heat required to melt the ice at 0°C. The heat of fusion of ice will be 334 J/g, so;

Q₂ = 10.0 g × 334 J/g = 3340 J

Heating the water from 0°C to 100°C

Now that all the ice has melted, we need to heat the resulting water from 0°C to its boiling point at 100°C. The specific heat capacity of water is 4.18 J/g°C, so;

Q₃ = 10.0 g × 4.18 J/g°C × (100°C - 0°C) = 4180 J

Vaporizing the water at 100°C

Once the water reaches its boiling point at 100°C, we need to vaporize it into steam. The heat of vaporization of water will be 40.7 kJ/mol, or 2260 J/g. Since we know that 18.0 g of water make up one mole, we can calculate the heat required to vaporize 10.0 g of water as;

Q₄ = 10.0 g × 2260 J/g = 22,600 J

Heating the steam from 100°C to 110°C

Finally, we need to heat the steam from 100°C to its final temperature of 110°C. The specific heat capacity of steam is 1.84 J/g°C, so;

Q₅ = 10.0 g × 1.84 J/g°C × (110°C - 100°C) = 184 J

Total heat required

To find the total heat required to warm the ice from -10.0°C to steam at 110.0°C, we simply add up all the heats calculated in the previous steps;

[tex]Q_{total}[/tex] = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

= 209 J + 3340 J + 4180 J + 22,600 J + 184 J

= 29,513 J

Therefore, the amount of heat is 29,513 J or 29.5 kJ.

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Answer:

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