Chapter 12.2, Problem 19PP

To determine

The amount of thermal energy to be added to a substance for a given change in temperature.

Answer to Problem 19PP

The amount of thermal energy to be added is $3.75\times {10}^4\text{ J}$

Explanation of Solution

Given:

The temperature of $1.00\times {10}^2 \text{g}$ of ice is changed from $-20°\text{C}$ to water at $0°\text{C}$ by adding a specific amount of thermal energy.

Formula Used:

When temperature of an object is changed from $T_A$ (in Kelvin) to $T_B$ , there is always a gain or loss of thermal energy.

Here a state change, melting of ice is involved and hence the thermal energy involved in this change has two parts-one being the energy involved in melting of solid and the energy for temperature change.

The energy for melting of ice of mass m is given by,

  $Q_{melt,ice}=m{H}_{\text{f}}$

In the above equation, the parameter $H_{\text{f}}$ represents the heat of fusion of ice.

The energy for causing temperature change is dependent on the temperature change ,mass of the object $m$ and specific heat $C$ , and is given by the expression,

  $Q_t=mC\left(T_B-T_A\right)$

Thus, the total thermal energy for this given conversion is ,

  $\begin{array}{c}Q=Q_t+Q_{melt,ice}\\ =mC\left(T_B-T_A\right)+m{H}_{\text{f}}\end{array}$ ...... (1)

Calculation:

Here, ice is having a mass of $m\text{=1}.00\times \text{1}{0}^{\text{2}}\text{g}$ . The temperature of ice is changed from

  $-20°\text{C}$ to water at $0°\text{C}$ . The specific heat of ice is $C=2060\text{ J/kg}\cdot \text{K}$ . The heat of fusion of ice is $H_{\text{f}}=3.34\times {10}^5\text{ J/kg}$

The initial temperature of ice in Kelvin can be calculated as,

  $\begin{array}{c}{T}_A=-20+273\\ =253\text{ K}\end{array}$

Similarly, the final temperature of water in Kelvin is,

  $\begin{array}{c}{T}_B=0+273\\ =273\text{ K}\end{array}$

Substituting these parameters in (1), the thermal energy to be added to achieve this temperature change is,

  $\begin{array}{c}Q=0.100\times 2060\left(273-253\right)+\left(0.100\right)\left(3.34\times {10}^5\right)\\ =3.75\times {10}^4\text{ J}\end{array}$

Conclusion:

Thus, the thermal energy required to raise the temperature of ice for the given range is $3.75\times {10}^4\text{ J}$ .