How much energy is absorbed when 10g of ice at 0degree Celsius becomes steam at 100 degree Celsius?

To find the amount of energy absorbed during the phase change from ice at 0 degrees Celsius to steam at 100 degrees Celsius, we need to consider the energy required for heating the ice, melting the ice, heating the water, and boiling the water.

First, let's calculate the energy required to heat the ice from 0°C to its melting point. The specific heat capacity of ice is 2.09 J/g°C, and the temperature change is 0°C. Therefore, the energy absorbed is:

Energy_1 = mass × specific heat capacity × temperature change
= 10g × 2.09 J/g°C × 0°C
= 0 J

Next, let's calculate the energy required to melt the ice at its melting point of 0°C. The heat of fusion for ice is 334 J/g. Therefore, the energy absorbed is:

Energy_2 = mass × heat of fusion
= 10g × 334 J/g
= 3340 J

After the ice has melted, we need to calculate the energy required to heat the water from 0°C to its boiling point. The specific heat capacity of water is 4.18 J/g°C, and the temperature change is 100°C. Therefore, the energy absorbed is:

Energy_3 = mass × specific heat capacity × temperature change
= 10g × 4.18 J/g°C × 100°C
= 4180 J

Finally, we need to calculate the energy required to convert the water into steam at its boiling point. The heat of vaporization for water is 2260 J/g. Therefore, the energy absorbed is:

Energy_4 = mass × heat of vaporization
= 10g × 2260 J/g
= 22600 J

To find the total energy absorbed, we sum up all the energies:

Total Energy = Energy_1 + Energy_2 + Energy_3 + Energy_4
= 0 J + 3340 J + 4180 J + 22600 J
= 30120 J

Therefore, the total energy absorbed is 30120 Joules.