Chapter 10, Problem 48A

Interpretation Introduction

Interpretation: The amount of energy (in calories) required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ needs to be calculated.

Concept introduction: Specific heat of capacity can be defined as the amount of heat added to a substance to raise the temperature by one degree.

Answer to Problem 48A

The amount of energy required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ is $\text{7641}\text{.5 cal}$ .

Explanation of Solution

The amount of energy (in calories) required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ can be calculated using formula:

  $\text{q = mc}\Delta \text{T}$

Where q is heat, m is mass, c is specific heat and $\Delta \text{T}$ is change in temperature.

The specific heat capacity for the liquid water is $4.184\text{J/g°C}$

Substituting the values:

  $\begin{array}{l}\text{q = 145 g×4}\text{.184 J/g°C×}\left(\text{75}\text{.0-22}\text{.3}\right)\text{°C}\\ \text{q = 606}\text{.68 J/g°C×}\left(\text{52}\text{.7}\right)\text{°C}\\ \text{q = 31972}\text{.036 J}\end{array}$

Now, converting the amount from J to cal as:

  $1\text{ cal = 4}\text{.184 J}$

So,

  $\begin{array}{l}\text{q = 31972}\text{.036 J×}\frac{\text{1 cal}}{\text{4}\text{.184 J}}\\ \text{q = 7641}\text{.5 cal}\end{array}$

Hence, the amount of energy required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ is $\text{7641}\text{.5 cal}$ .

Conclusion

  $\text{7641}\text{.5 cal}$ is the amount of energy required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ .