Chapter 17, Problem 91A

a.

Interpretation Introduction

Interpretation: To calculate the amount of heat required when the ice cube absorbs from the water when it melts.

Concept Introduction: The heat that is needed to raise the temperature by $1°\text{C}$ of any $1\text{g}$ of substance is known as specific heat.

It can be expressed as follows:

  $\text{Specific heat}=\frac{\text{Heat}}{\text{Mass}\times \text{ΔT}}$

Answer to Problem 91A

The amount of heat required is $13,378\text{ J}$ .

Explanation of Solution

The value for the heat of fusion is $\text{334}\text{.45 J/g}$ .

The mass of ice is $40.0\text{g}$ .

Therefore, the amount of heat can be calculated as follows:

  $\begin{array}{c}q=40.0\cancel{\text{g}}\times \text{334}\text{.45 J/}\cancel{\text{g}}\\ \text{=13,378 J}\end{array}$

b.

Interpretation Introduction

Interpretation: To calculate the grams of water needed.

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 91A

The number of grams of water are $127.9\text{ g}$ .

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 change in temperature, $m$ represents the mass of a substance, and $C$ represents the specific heat.

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

The amount of heat required is $13,378\text{ J}$ .

The change in temperature is given as:

  $\begin{array}{c}\Delta T=25\text{°C}-0\text{°C}\\ \text{=25°C}\end{array}$

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

  $\begin{array}{c}q=\Delta H=m\times C\times \Delta T\\ 13,378\cancel{\text{J}}=m\times \text{4}\text{.18 }\cancel{\text{J}}\text{/}\left(\text{ g}\cancel{\text{°C}}\right)\times 25\cancel{\text{°C}}\\ m=127.9\text{g}\end{array}$

Thus, the mass of water in grams is $127.9\text{ g}$ .