Chapter U5, Problem 5RE

(a)

Interpretation Introduction

Interpretation: The amount of calories of energy transferred to raise the temperature of 50 g of water from 25 ^oC to boiling needs to be determined.

Concept Introduction:

The amount of energy transferred for the given mass of substance can be calculated as follows:

  $q=mc\Delta T$

Here, m is mass, c is specific heat capacity and $\Delta T$ is change in temperature.

Answer to Problem 5RE

3750 cal.

Explanation of Solution

The specific heat capacity of water is 1 cal/g ^oC. The mass of water is 50 g. Also, the initial and final temperature is 25 ^oC and 100 ^oC respectively.

The calories of energy can be calculated as follows:

  $q=mc\Delta T$

Putting the values,

  $q=\left(50\text{ g}\right)\left(\text{1}{\text{.0 cal/g }}^{\text{o}}\text{C}\right)\left(\text{100}-{25}^{\text{o}}\text{C}\right)=3750\text{ cal}$

The amount of calories of energy transferred to raise the temperature of 50 g of water from 25 ^oC to boiling is 3750 cal.

(b)

Interpretation Introduction

Interpretation: The amount of calories of energy transferred to lower the temperature of 100 g of water from boiling to 25 ^oC needs to be determined.

Concept Introduction:

The amount of energy transferred for the given mass of substance can be calculated as follows:

  $q=mc\Delta T$

Here, m is mass, c is specific heat capacity and $\Delta T$ is change in temperature.

Answer to Problem 5RE

-7500 cal.

Explanation of Solution

The specific heat capacity of water is 1 cal/g ^oC. The mass of water is 100 g. Also, the initial and final temperature is 100 ^oC and 25 ^oC respectively.

The calories of energy can be calculated as follows:

  $q=mc\Delta T$

Putting the values,

  $q=\left(100\text{ g}\right)\left(\text{1}{\text{.0 cal/g }}^{\text{o}}\text{C}\right)\left({\left(25-100\right)}^{\text{o}}\text{C}\right)=-7500\text{ cal}$

Thus, the amount of calories of energy transferred to lower the temperature of 100 g of water from boiling to 25 ^oC is -7500 cal.