Heat required to melt an iceberg

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The discussion centers on calculating the heat required to melt 12 percent of a 1.30×10^5 metric-ton iceberg, emphasizing the need for the correct formula. Participants clarify that the initial equation used, Q = c*m*change in temperature, is inappropriate for a phase change, which does not involve a temperature change. Instead, the correct approach involves using the latent heat of fusion, Q = m L_f, where L_f is approximately 80 kcal/kg for water. This distinction is crucial for accurately determining the energy needed for melting the iceberg. Understanding these thermodynamic principles is essential for solving the problem correctly.
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Homework Statement



I think I'm doing this problem correctly... I'm not sure about my change in temperature value. Any suggestions?



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 12 percent of a 1.30×105 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.)

Homework Equations



Q = c*m*change in temperature

The Attempt at a Solution




Miceberg = 1.30E8 kg
12% of iceberg = 1.56E7 kg
C ice = 2.00E3

Q = 2.00E3 * 1.30E8 * 1

Q = 3.12 E 5 J
 
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Ah... thank you
 
Youre using the wrong equation. Like has been stated, its a phase change and the formula would be Q = m L_{f} where L_{f} is the latent heat of fusion for water. IIRC its around 80 kcal/kg for water.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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