# HELP! Melting Iceberg: Latent heat of fusion

URGENT HELP! Melting Iceberg: Latent heat of fusion

## Homework Statement

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 15 percent of a 3.90×10^5 metric-ton iceberg? One metric ton is equal to 10^3 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 6.0×10^13 J.)

Q = mLf

## The Attempt at a Solution

first I converted 3.90 x10^5 metric tons to 3.90 x 10^9 kg since one metric tons is equal to 10^3 kg. Then I multiplied the 3.90 x 10^9 kg times .15 to get 15% of the iceberg mass and got 5.85 x 10^8 kg. Lastly I multiplied 5.85 x 10^8 times the latent heat of fusion of water (334000 J/kg) and got 1.95 x 10^14 J but that's incorrect. I'm pretty sure I'm doing it right but I'm not getting the right answer can anyone help?

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Astronuc
Staff Emeritus
3.90 x10^5 metric tons to 3.90 x 10^9 kg
should be 3.8 E8 kg, and .15 of that is 5.85 E7 kg

1 MT = 1000 kg.

the 334 kJ/kg is correct.

One should be getting 1.954 E13 J.

thanks a bunch!