0.1*4200*(26-11) = (0.03*lf) + (0.03*4200*11)
6300=0.03lf + 737.2
Lf=5560.8/0.03 = 3.4755 * 25JKg-1
Moraa orina answered the question on April 10, 2018 at 18:00
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A block of ice of mass 40g at 0^oC is placed in a calorimeter containing 400g of water at 20^oC. Ignoring heat absorbed by the...
(Solved)
A block of ice of mass 40g at 0^oC is placed in a calorimeter containing 400g of water at 20^oC. Ignoring heat
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Date posted:
April 10, 2018
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April 10, 2018
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Date posted:
April 10, 2018
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Answers (1)
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April 10, 2018
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April 10, 2018
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April 10, 2018
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April 10, 2018
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