Sunday

December 21, 2014

December 21, 2014

Posted by **Lindsay** on Tuesday, May 15, 2007 at 10:24pm.

I take 1.60×10^5 x 103 x 8/100 x 1000 x 333 (heat of fusion of ice) and I keep getting the wrong answer. What am I doing wrong?

4.40E11? I used 334 instead of 333.

Yep, and it still says it's wrong!

Am I doing everything else right?

1.60E5 mtons*(103 kg/mton)*(1000 g/kg)*0.08*334 J/g = 4.40E11 Joules.

Check your numbers in the problem with those in the post. Check the units to see they want the answer in J and not kJ or some other unit. Check that the conversion factor is correct. I went to Google and typed in 1 metric ton and it tells me 1 metric ton = 1,000 kg. That's about a factor of 10 different.

I checked google again. Typed in define metric ton and there are a "ton" of returns I obtained, each saying 1 metric ton = 1,000 kg. That isn't the number in your post.

Also, I would be interested in knowing the value of your answer to the 12.1 g ice problem.

I think I just figured out what is wrong. Of course we know it's the unit BUT I'll bet that the 103 was meant to be 10^3 which is 1000 and 1000 is what I found as the conversion factor. Try the 1.6E5*1000*1000*0.08*334=4.27E12 Joules.

Ok I got it. I erased the 103 from the problem and put in 1000 instead. This gave me the number of 4.3E12, which was correct. Thanks for correcting that for me...it WAS typed like that in the problem on my h/w, for some reason. It's obviously not right.

My answer to the last problem I posted was 0.73 J/gK.

OH I see! 10^3! They didn't have it typed like that on the problem, it just looked like 103. But it makes more sense now.

- Chemistry -
**john**, Wednesday, February 11, 2009 at 6:02pmIcebergs 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 14 percent of a 2.40×105 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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