Chapter U5.10, Problem 4E

(a)

Interpretation Introduction

Interpretation: The energy transferred to water on combustion of a given amount of propanol is to be calculated.

Concept introduction: Specific heat is defined as quantity of heat required to raise the temperature of one gram substance by one Kelvin. It is represented by ‘c’.

Answer to Problem 4E

The heat absorbed is $100cal$

Explanation of Solution

The quantity of heat can be calculated by using formula:

  $\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatabsorbed\\ m=givenmass\\ \Delta T=riseintemperature\\ cisspecificheat\end{array}$

Given data:

  $\begin{array}{c}m=4g\\ \Delta T=45-20\\ =25°C\\ c=1cal/g-°C\end{array}$

Substituting the values in formula:

  $\begin{array}{c}q=mc\Delta T\\ =4\times 1\times 25\\ =100cal\end{array}$

(b)

Interpretation Introduction

Interpretation: The amount of kilocalories transferred in a process that involves combustion of propanol is to be calculated.

Concept introduction: The given amount of energy is to be converted into kilocalories.

The conversion factor is:

  $1kcal=1000cal$

Answer to Problem 4E

  $100cal=0.1kcal$

Explanation of Solution

As, $1kcal=1000cal$

Therefore,

  $\begin{array}{l}100cal=100cal\times \frac{1kcal}{1000cal}\\ =0.1kilocalories\end{array}$

(c)

Interpretation Introduction

Interpretation: The kilojoules of energy transferred in a process that involves combustion of propanol needs to be determined.

Concept introduction: The given amount of energy is to be converted into kilocalories.

The conversion factor is:

  $1kcal=1000cal$

  $1kcal=4.184kJ$

Answer to Problem 4E

  $100cal=0.4184kJ$

Explanation of Solution

As, $1kcal=1000cal$

  $1kcal=4.184kJ$

Therefore,

  $\begin{array}{l}100cal=100cal\times \frac{1kcal}{1000cal}\times \frac{4.184kJ}{1kcal}\\ =0.4184kJ\end{array}$

(d)

Interpretation Introduction

Interpretation: The temperature increase for $300mL$ water is to be calculated.

Concept introduction:

Specific heat is defined as quantity of heat required to raise the temperature of one gram substance by one Kelvin. It is represented by ‘c’.

The quantity of heat can be calculated by using formula:

  $\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatabsorbed\\ m=givenmass\\ \Delta T=riseintemperature\\ cisspecificheat\end{array}$

Answer to Problem 4E

There will same increase in temperature on increasing the volume of water.

Explanation of Solution

Change in temperature depends on quantity of substance which is added to water. So on changing the volume of water, temperature change will remain same.

(e)

Interpretation Introduction

Interpretation: The temperature increase for $2g$ propanol is to be calculated.

Concept introduction:

Specific heat is defined as quantity of heat required to raise the temperature of one gram substance by one Kelvin. It is represented by ‘c’.

The quantity of heat can be calculated by using formula:

  $\begin{array}{l}q=mc\Delta T\\ whereqisquantityofheatabsorbed\\ m=givenmass\\ \Delta T=riseintemperature\\ cisspecificheat\end{array}$

Answer to Problem 4E

Increase in temperature will become double when mass of substance is reduced to half of its value.

Explanation of Solution

According to the formula, increase in temperature is inversely proportional to the mass of substance added. Therefore, increase in temperature will become double when mass of propanol is decreased from $4g$ to $2g$

This can be represented as follows:

  $\begin{array}{c}q=mc\Delta T\\ 100\text{ cal}=2\times 1\times \Delta T\\ \Delta T=\frac{100}{2}\\ =50\end{array}$