Question

(II) How long does it take the Sun to melt a block of ice at 0°C with a flat horizontal area \({\bf{1}}{\bf{.0}}\;{{\bf{m}}^{\bf{2}}}\) and thickness 1.0 cm? Assume that the Sun’s rays make an angle of 35° with the vertical and that the emissivity of ice is 0.050

Short answer

The time taken to melt the ice is \(20.7\;{\rm{h}}\).

Step-by-step solution

Understanding the radiation process

Radiation is the process that does not need any medium to carry heat from one place to another. It involves energy transfer by electromagnetic waves, such as from the sun.

Given data

The area of the ice block is \(A = 1.0\;{{\rm{m}}^{\rm{2}}}\).

The thickness of the ice block is \(x = 1.0\;{\rm{cm}}\).

The emissivity of the ice is \(\varepsilon = 0.050\).

The angle made by sunrays with vertical is \(\theta = 35^\circ \).

Evaluation of the time taken to melt the ice

The expression for the rate at which the heat energy absorbed by the ice to melt can be written as:

\(\frac{Q}{t} = S\varepsilon A\cos \theta \)

Here, \(S\) is the intensity of the sunlight and its value is \(S = 1000\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}}}\).

Rewrite the above equation as follows:

\(t = \frac{Q}{{S\varepsilon A\cos \theta }}\) … (i)

The expression for the amount of heat absorbed by the ice can be written as:

\(Q = m{L_{\rm{f}}}\) … (ii)

Here, \({L_{\rm{f}}}\) is the latent heat of fusion of ice, and its value is \(3.33 \times {10^5}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}}\).

The expression for the mass of ice can be written as:

\(m = \rho Ax\) … (iii)

Here, \(\rho \) is the density of the ice, and its value is \(917\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.\\} {{{\rm{m}}^{\rm{3}}}}}\).

Substitute the value of equation (iii) in equation (ii).

\(Q = \rho Ax{L_{\rm{f}}}\) … (iv)

Substitute the value of equation (iv) in equation (i).

\(\begin{array}{c}t = \frac{{\rho Ax{L_{\rm{f}}}}}{{S\varepsilon A\cos \theta }}\\t = \frac{{\rho x{L_{\rm{f}}}}}{{S\varepsilon \cos \theta }}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}t = \frac{{\left( {917\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.\\} {{{\rm{m}}^{\rm{3}}}}}} \right)\left[ {\left( {1\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right]\left( {3.33 \times {{10}^5}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}}} \right)}}{{\left( {1000\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}}}} \right)\left( {0.050} \right)\cos \left( {35^\circ } \right)}}\\ = 7.5 \times {10^4}\;{\rm{s}}\left( {\frac{{{\rm{1}}\;{\rm{h}}}}{{{\rm{3600}}\;{\rm{s}}}}} \right)\\ \approx 20.7\;{\rm{h}}\end{array}\)

Thus, the time taken to melt the ice is \(20.7\;{\rm{h}}\).