Chapter U5, Problem 9STP

Interpretation Introduction

Interpretation:

The phase change versus time of a mass of ice as heat is applied to the ice at a constant rate is shown. The state of water at a particular temperature of the plot needs to be identified.

Concept introduction:

As heat is applied to a mass of solid ice kept at sub-zero temperature, the temperature rises gradually till the normal melting point of ice (0ºC at atmospheric pressure) is reached. At this point solid ice changes completely to liquid water. There-after application of heat raises the temperature of liquid water from 0ºC to 100ºC which is the normal boiling point of liquid water. At the normal boiling point of water, a second phase change occurs and water coverts to the gaseous state known as steam. Further, addition of heat increases the temperature of steam above 100ºC.

Answer to Problem 9STP

The second option, option (B) is the correct answer.

Explanation of Solution

Reason for Correct Option:

The normal boiling point of liquid water is 100ºC and at this temperature, liquid water converts spontaneously to steam at 1 atmosphere pressure. Since the time D in the heating curve of ice indicates a constant temperature of 100ºC, hence, both liquid water and water vapor or steam is present at time D. This is because at time D, some water is converted to water vapor and further heating of the sample is required till the phase change is completed, i.e., all the water is converted to water vapor.

Conclusion

Reasons for Incorrect Options:

Since the temperature at time D is much higher than the normal melting point of ice, hence, ice cannot be present at time D. Thus, option (A) is incorrect.

Further, water vapor is the only component in a system when the temperature is above 100ºC; however, as the system is maintained at 100ºC at time D, hence, both options (C) and (D) are incorrect.