How many grams of ice at -23°C must be added to 176 grams of water that is initi

Question

How many grams of ice at -23°C must be added to 176 grams of water that is initially at a temperature of 69 degrees c, to produce water at a final temperature 8degrees c of Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 j/kg and of ice is 2000 j/kg. For water the normal melting point is 0 degrees c, and the heat of fusion is 334 x 10^3 j/kg The normal boiling point is 100 degrees c, and the heat of vaporization is 2.256 x10^6 j/kg

Explanation / Answer

Heat gained by ice = Heat lost by water Heat gained by ice = Heat to warm ice + heat of fusion + heat to warm melted ice heat needed to warm = m*c * change in temperature = micecice(0-(-23)) + mcfusion + micecwater(8- 0)= 23 micecice + micecfusion + 8micecwater mice(23cice + cfusion + 8cwater) Heat lost by water : mwater*cwater*decrease in temperature = 0.176(4190)(69-8) = 737.44(61)=44983.84 mice(23(2000) + 334,000 + 8(4190)) = 413520 mice = 44983.84 mice = 0.10878 kg = 108.78 g

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