Chapter 10, Problem 49A

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 49A

The amount of energy (in calories) to heat $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ is $7.64\times {10}^3\text{cal}$ .

Explanation of Solution

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

The amount of heat required for heating $\text{145}\text{g}$ of water from $22.3°\text{C}$ to $75.0°\text{C}$ can be calculated as:

  $\begin{array}{c}Q=s\times m\times \Delta T\\ =4.184\text{J/g°C}\times 145\text{g}\times \left(75.0-22.3\right)\text{°C}\\ \text{=31972}\text{J}\times \frac{\text{1}\text{cal}}{\text{4}\text{.184}\text{J}}\\ =7.64\times {10}^3\text{cal}\end{array}$

Conclusion

The equation for heat is as follows:

  $Q=s\times m\times \Delta T$

Where, Q is the amount of heat, m is the mass of substance, s is the specific heat capacity and $\Delta T$ is the temperature change.