Question

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Answer 1

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ