Chapter 17, Problem 51A

(a)

Interpretation Introduction

Interpretation:

The given conversion is to be made.

Concept Introduction : A calorie is defined as the quantity of heat needed to raise the temperature of $\text{1 g}$ of pure water by $\text{1°C}$ . It is represented by a small c or cal.

Calorie refers to the energy in food. It is represented by C.

Answer to Problem 51A

  $\text{0}\text{.85 Calorie}$

Explanation of Solution

As we know 1 Calorie = 1000 calories

Thus 1 calorie = 0.001 Calories

  $\text{8}\text{.50}\times {\text{10}}^2\text{ cal}$ = $\frac{\text{0}\text{.001 Calorie}}{\text{1}\text{.0 cal}}\text{×8}\text{.50}\times {\text{10}}^2\text{ cal=0}\text{.85 Calorie}$

(b)

Interpretation Introduction

Interpretation:

The given conversion is to be made.

Concept Introduction :

A calorie is defined as the quantity of heat required to raise the temperature of $\text{1 g}$ of pure water by $\text{1°C}$ . It is represented by a small c or cal.

A joule of heat is defined as the quantity of heat required to raise the temperature of $\text{1 g}$ of pure water by $\text{0}\text{.2390°C}$ .

Answer to Problem 51A

  $\text{1858 Joule}$

Explanation of Solution

As we know $\text{1 calorie = 4}\text{.184 Joules}$

Thus $\text{444 calorie}$ = $\frac{4.184\text{ Joule}}{\text{1}\text{.0 cal}}\text{×444 cal=1858 Joule}$

(c)

Interpretation Introduction

Interpretation:

The given conversion is to be made.

Concept Introduction : A kilojoule of heat is defined as the quantity of heat required to raise the temperature of $\text{1}\text{.0 kg}$ of pure water by $\text{0}\text{.2390°C}$ .

A joule of heat is defined as the quantity of heat required to raise the temperature of $\text{1 g}$ of pure water by $\text{0}\text{.2390°C}$ . It is represented by J.

Answer to Problem 51A

  $\text{1880 Joules}$

Explanation of Solution

As we know $\text{1 kJ = 1000 Joules}$

Thus $\text{1}\text{.88 kJ = }\frac{\text{1000 Joules}}{\text{1}\text{.0 kJ}}\text{×1}\text{.88 kJ=1880 Joules}$

(d)

Interpretation Introduction

Interpretation:

The given conversion is to be made.

Concept Introduction:

A kilojoule of heat is defined as the quantity of heat required to raise the temperature of $\text{1}\text{.0 kg}$ of pure water by $\text{0}\text{.2390°C}$ .

A calorie is defined as the quantity of heat required to raise the temperature of $\text{1 g}$ of pure water by $\text{1°C}$ . It is represented by a small c or cal.

Answer to Problem 51A

  $\text{107}\text{.5 cal}$

Explanation of Solution

As we know, $\text{1 kJ = 239}\text{.006 cal}$

Thus $\text{ 4}\text{.5}\times {\text{10}}^{-1}\text{ kJ = }\frac{239.006\text{ cal}}{\text{1}\text{.0 kJ}}\text{×4}\text{.5}\times {\text{10}}^{-1}\text{ kJ =107}\text{.5 cal}$