Answer:
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.
Step-by-step 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