Chapter 12, Problem 6STP

To determine

To Choose: The correct option.

Answer to Problem 6STP

Option C

Explanation of Solution

Given:

Mass of ice, $m=87\text{g}$

Initial temperature of ice, $T_1=14\text{}\text{K}$

Final temperature, $T_2=140°\text{C}$

Formula Used:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

  $0°\text{C}=273\text{K}$

Calculations:

Convert initial temperature of ice:

  $\begin{array}{c}{T}_1=\left(14-273\right)°\text{C}\\ =-259°\text{C}\end{array}$

The temperature of ice would rise from $-259°\text{C}$ to $0°\text{C}$ . Then the ice would melt. Then the temperature of water would rise from $0°\text{C}$ to $100°\text{C}$ . At $100°\text{C}$ , water would vaporize. Then temperature of steam would rise from $100°\text{C}$ to $140°\text{C}$ .

Heat required to raise the temperature of ice from $-259°\text{C}$ to $0°\text{C}$ can be calculated as follows:

Specific heat of ice, $c_i=2.03\text{J/g°C}$

  $\begin{array}{c}{Q}_1=m{c}_i\Delta T\\ =\left(87\text{g}\right)\left(2.03\text{J/g°C}\right)\left(0°\text{C}-\left(-259°\text{C}\right)\right)\\ =45741.9\text{J}\end{array}$

Latent heat of fusion of ice, $L_f=334\text{J/g}$

Heat required to melt the ice:

  $\begin{array}{c}{Q}_2=m{L}_f\\ =87\text{g}\times \text{334J/g}\\ \text{=29058}\text{J}\end{array}$

Heat required to raise the temperature from $0°\text{C}$ to $100°\text{C}$ .

Specific heat capacity of water is, $c_w=4.184\text{J/g}\cdot \text{°C}$

  $\begin{array}{c}{Q}_3=m{c}_w\Delta T\\ =\left(87\text{g}\right)\left(4.184\text{J/g}\cdot \text{°C}\right)\left(100°\text{C-0}°\text{C}\right)\\ =36400.8\text{J}\end{array}$

Latent heat of vaporization of water, $L_v=2260\text{J/g}$

Heat required to vaporise the water:

  $\begin{array}{c}{Q}_4=m{L}_v\\ =87\text{g}\times 2260\text{J/g}\\ \text{=196620}\text{J}\end{array}$

Specific heat of steam, $c_s=1.99\text{J/g}\cdot °\text{C}$

Heat required to raise the temperature from $100°\text{C}$ to $140°\text{C}$

  $\begin{array}{c}{Q}_5=m{c}_s\Delta T\\ =\left(87\text{g}\right)\left(1.99\text{J/g}\cdot °\text{C}\right)\left(40°\text{C}\right)\\ =6925.2\text{J}\end{array}$

Total energy required:

  $\begin{array}{c}Q=Q_1+Q_2+Q_3+Q_4+Q_5\\ =45741.9\text{J}+2905\text{8}\text{J}+36400.8\text{J+196620}\text{J+6925}\text{.2 J}\\ =314767.9\text{J}\\ \approx \text{315 kJ}\end{array}$

Conclusion:

Thus, correct option is C.