Chapter 22, Problem 94A

Interpretation Introduction

Interpretation: To calculate the number of calories absorbed.

Concept Introduction: At constant pressure, the amount of heat that is released or absorbed is identical to the enthalpy change.

It can be expressed as follows:

  $q=\Delta H$

Here, $q$ represents the heat change and $\Delta H$ represents the enthalpy change.

Answer to Problem 94A

The amount of heat absorbed is $5594.6\text{ Cal}$ .

Explanation of Solution

The formula that is used to calculate the amount of heat absorbed or released can be expressed as follows:

  $q=\Delta H=m\times C\times \Delta T$ ……(1)

Here, $q$ represents the heat change, $\Delta H$ represents the enthalpy change, $\Delta T$ represents the temperature change, $m$ represents the mass of a substance, and $C$ represents the specific heat.

The mass of s substance is $56.0\text{ g}$ .

The specific heat of liquid water is $\text{4}\text{.18 J/}\left(\text{g}\text{.°C}\right)$ .

The temperature change is given as:

  $\begin{array}{c}\Delta T=\text{100°C}-\text{0°C}\\ \text{=100°C}\end{array}$

Now, substitute the above values in equation (1) in order to calculate the amount of heat absorbed.

  $\begin{array}{c}q=\Delta H=m\times C\times \Delta T\\ =56.0\cancel{\text{g}}\times \text{4}\text{.18 J/}\left(\cancel{\text{g}\text{.°C}}\right)\times \text{100}\cancel{\text{°C}}\\ =23408\text{ J}\end{array}$

Conclusion

Thus, the amount of heat absorbed is $5594.6\text{ Cal}$ .