Chapter U5, Problem SI8RE

(a)

Interpretation Introduction

Interpretation:

Heating curve of water must be drawn to show the different stages when water is heated from 10°C to 120°C.

Concept Introduction:

Heating curve is the graphical representation of phase change when a substance is heated.

Freezing point and boiling point of water is 0°C and 100°C respectively.

Answer to Problem SI8RE

Heating curve is shown below.

  

Explanation of Solution

In the heating curve it is shown that when water is heated from 10°C then at 100°C the phase changes from liquid water to water vapor and this is the boiling point. Then temperature is again increasing from 100°C to 120°C when further heat is given.

b)

Interpretation Introduction

Interpretation:

Amount of heat energy must be calculated which is transferred to water when the temperature is increased from 10°C to 120°C.

Concept Introduction:

Heat energy can be determined as follows:

  $\text{H=m×s×t}\text{.}$

Here m, s and t are mass, specific heat and temperature change respectively.

Answer to Problem SI8RE

Heat energy is 5500 cal.

Explanation of Solution

Initial temperature of water is 10°C and final temperature is 120°C. So temperature increase t=110°C. Specific heat(s) of water is 1cal/g.

Mass of the given water is 50 g.

Thus heat required to increase the temperature from 10°C to 120°C can be calculated as follows:

  $\begin{array}{l}\text{H=m×s×t=50g×1cal/g°C×110°C}\\ \text{=5500 cal}\text{.}\end{array}$

c)

Interpretation Introduction

Interpretation:

Which part of the heating process involves most heat transfer must be found from the calculation and heating curve.

Concept Introduction:

Heat energy required for the increase of temperature (t) of mass in grams of any substance with specific heat s can be written as follow:

  $\text{H=m×s×t}\text{.}$

Answer to Problem SI8RE

Heating of water from 10°C to 100°C involves most heat transfer.

Explanation of Solution

  

From the heating curve, it is clear that liquid water portion needs more heat as the graph is steeply rising.

The heat energy required for the increase of temperature of 50 g water from 10°C to 100°C

  $\begin{array}{l}\text{H=m×s×t=50g×1cal/g°C×100°C}\\ \text{=5000 cal}\text{.}\end{array}$

The heat energy required for the increase of temperature of 50 g water vapor from 100°C to 120°C

  $\begin{array}{l}\text{H=m×s×t=50g×0}\text{.5cal/g°C×100°C}\\ \text{=2500 cal}\text{.}\end{array}$