Calculate the amount of heat required to convert 10.0 grams of ice at –20.°C to steam at 120.°C. (Sp. heat of H2O(s) = 2.09 J/g•°C, Sp. heat of H2O(l) = 4.18 J/g•°C, Sp. heat of H2O(gas) = 2.03 J/g•°C; heat of fusion of H2O(solid) = 333 J/g, heat of vaporization of H2O(liquid) = 2260 J/g).

THE AMOUNT OF HEAT REQUIRED TO CONVERT ICE FROM -20 C TO STEAM AT 120 C IS 30 946 J OR 30.946 KJ OF HEAT.

Explanation:

Mass = 10 g

To convert 10 g of ice at -20°C to steam at 120°C, the heat involved is:

1. Heat involved in converting the ice from -20 °c to ice at 0 °C:

Heat = mass * specific heat of water solid * change in temperature

heat = 10g * 2.09 J/g°C * ( 0- (-20))

Heat = 10 * 2.09 * 20

heat = 418 J

2. Heat required to convert the ice from 0°C to water at 0°C:

Heat = mass * specific heat of fusion of water solid

Heat = 10 * 333

Heat = 3330 J

3. Heat required to convert water at 0 C to water at 100 C:

Heat = mass * specific heat of water * change in teperature

Heat = 10 * 4.18 * (100 -0)

Heat = 4180 J

4. Heat required to convert water at 100 C to steam at 100 C:

Heat = mass * specific heat of vaporization

Heat = 10 * 2260

Heat = 22600 J

5. Heat required to convert steam from 100 C to steam at 120 C:

Heat = mass * specific heat of water * change in temperature

Heat = 10 * 2.09 * (120 -100)

Heat = 10 * 2.09 * 20

Heat = 418 J

T

he heat required to convert 10 g of ice at -20 C to steam at 120 C is therefore the total of the individual heat of reactions

Total amount of heat = ( 418 J + 3330 J + 4180 J + 22600 J + 418 J)

Total heat =  30946 J