Icebergs in the North Atlantic present hazards to shipping, causing the length of shipping routes to increase by about 30 percent during the iceberg season. Attempts to destroy icebergs include planting explosives, bombing, torpedoing, shelling, ramming, and painting with lampblack. Suppose that direct melting of the iceberg, by placing heat sources in the ice, is tried. How much heat is required to melt 9 percent of a 2.50×105 metric-ton iceberg? One metric ton is equal to 103 kg. Assume that the iceberg is at 0°C. (Note: To appreciate the magnitude of this energy, compare your answer to the Hiroshima atomic bomb which had an energy equivalent to about 15,000 tons of TNT, representing an energy of about 6.0×1013 J.)
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To determine the amount of heat required to melt 9 percent of a 2.50×10^5 metric-ton iceberg, we can use the formula:
Q = m * L
where
Q is the heat required,
m is the mass of the iceberg, and
L is the latent heat of fusion.
First, let's convert the mass of the iceberg to kilograms:
1 metric ton = 10^3 kg
Given that the mass of the iceberg is 2.50×10^5 metric tons, we can convert it to kilograms:
mass = 2.50×10^5 * 10^3 = 2.50×10^8 kg
Next, we need to find the latent heat of fusion. The latent heat of fusion is the amount of heat energy required to change a substance from a solid to a liquid at its melting point. For ice, the latent heat of fusion is approximately 3.34×10^5 J/kg.
Now we can calculate the heat required:
Q = 0.09 * mass * L
Substituting the values:
Q = 0.09 * 2.50×10^8 kg * 3.34×10^5 J/kg
Multiply the numbers:
Q = 0.09 * 2.50×10^8 * 3.34×10^5 J
Q ≈ 7.4545×10^12 J
Therefore, approximately 7.4545×10^12 joules of heat energy are required to melt 9 percent of the iceberg.
As a comparison, the energy equivalent of the Hiroshima atomic bomb is about 6.0×10^13 J. Comparing this to the heat required to melt 9 percent of the iceberg, we can see that the magnitude of energy required to melt the iceberg is significantly lower.