Chapter 14, Problem 123QAP

To determine

(a)

The amount of heat required to melt the ice.

Answer to Problem 123QAP

The amount of heat required to melt the ice is $7.1\times {10}^{20}\text{J}$.

Explanation of Solution

Given info:

Area covered by melted ice is, $A=720000{\text{km}}^{\text{2}}$.

Thickness of melted ice is, $l=3.0\text{m}$.

Initial temperature of ice is, $T_{\text{i}}=-10°\text{C}$.

Formula used:

Formula for the heat required to change the temperature is,

  $Q=mc\Delta T$

Formula for the heat required to change the phase is,

  $Q=mL$

Calculation:

Density of melted ice is,

  $\begin{array}{c}{\rho }_{\text{ice}}=92\%\times {\rho }_{\text{w}}\\ {\rho }_{\text{ice}}=\frac{92}{100}\times 1000{\text{kg/m}}^{\text{3}}\\ {\rho }_{\text{ice}}=920{\text{kg/m}}^{\text{3}}\end{array}$

The mass of the melted ice can be calculated as,

  $\begin{array}{c}m={\rho }_{\text{ice}}\times V\\ m={\rho }_{\text{ice}}\times \left(A\times l\right)\\ m=920{\text{kg/m}}^{\text{3}}\times \left(720000\times {10}^6{\text{m}}^{\text{2}}\times 3.0\text{m}\right)\\ m=19.872\times {10}^{14}\text{kg}\end{array}$

The amount of heat required to melt the ice is,

  $\begin{array}{l}Q=m{c}_{\text{ice}}\Delta T+m{L}_{\text{f}}\\ Q=19.872\times {10}^{14}\times 2093\times \left(10-0\right)+19.872\times {10}^{14}\times 334\times {10}^3\\ Q\approx 7.1\times {10}^{20}\text{J}\end{array}$

Conclusion:

Thus, the amount of heat required to melt the ice is $7.1\times {10}^{20}\text{J}$.

To determine

(b)

The amount of gasoline that contains as much energy as in part (a).

Answer to Problem 123QAP

The amount of gasoline that contains as much energy as in part (a) is $5.46\times {10}^{12}\text{gal}$.

Explanation of Solution

Given info:

The heat required to melt the ice is, $Q=7.1\times {10}^{20}\text{J}$.

Heat released by 1 gallon of gasoline is, $Q^{\prime }=1.3\times {10}^8\text{J/gal}$.

Formula used:

Formula to calculate the amount of gasoline is,

  $V=\frac{Q}{Q^{\prime }}$

Calculation:

Substituting the values in the above equation, we get

  $\begin{array}{c}V=\frac{7.1\times {10}^{20}\text{J}}{1.3\times {10}^8\text{J/gal}}\\ V=5.46\times {10}^{12}\text{gal}\end{array}$

Conclusion:

Thus, the amount of gasoline that contains as much energy as in part (a) is $5.46\times {10}^{12}\text{gal}$.

To determine

(c)

The amount of coal needed to be burned to produce the energy to melt the ice.

Answer to Problem 123QAP

The amount of coal needed to be burned to produce the energy to melt the ice is $3.3\times {10}^{10}\text{ton}$.

Explanation of Solution

Given info:

The heat required to melt the ice is, $Q=7.1\times {10}^{20}\text{J}$.

Heat produced by 1 ton of coal is, ${Q^{\prime }}^{\prime }=21.5\times {10}^9\text{J/ton}$.

Formula used:

Formula to calculate the amount of gasoline is,

  $m=\frac{Q}{{Q^{\prime }}^{\prime }}$

Calculation:

Substituting the values in the above equation, we get

  $\begin{array}{c}m=\frac{7.1\times {10}^{20}\text{J}}{21.5\times {10}^9\text{J/ton}}\\ m=3.3\times {10}^{10}\text{ton}\end{array}$

Conclusion:

Thus, the amount of coal needed to be burned to produce the energy to melt the ice is $3.3\times {10}^{10}\text{ton}$.