How much heat energy, in kilojoules, is required to convert 37.0g of ice at -18.

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Explanation / Answer

The total energy required is the " sum of the energies to heat the -18 °C ice to 0 °C ice and then melting the 0 °C ice into 0 °C water and further heating the water to 25 °C ". Ice at -18C ----------> Ice at 0C --------> Water at 0C ---------> Water at 25C Step 1: Heat required to raise the temperature of ice from -18 °C to 0 °C Use the formula Q = m * C * ?T where Q = heat energy m = mass of ice c = specific heat of ice =2.09 J/g·°C ?T = change in temperature Q1 = (37 g)x(2.09 J/g·°C)[(0 °C - -18 °C)] Q1 = (37 g)x(2.09 J/g·°C)x(18 °C) Q1 = 1391.94 J Heat required to raise the temperature of ice from -18 °C to 0 °C = 1391.94 J I would point out that ?T= T final -T initial. This is crucial as it would change the sign of the heat transfert. The signification of this is that, if it is a + sign, we GIVE energy to the system and , if it is a - sign, we REMOVE energy to the system Step 2: Heat required to convert 0 °C ice to 0 °C water Use the formula Q = m·?Hf where Q = heat energy m = mass ?Hf = heat of fusion=334 J/g Q 2= (37 g)x(334 J/g) Q 2= 12358 J Heat required to convert 0 °C ice to 0 °C water = 12358 J Step 3: Heat required to raise the temperature of 0 °C water to 25 °C water Q = m c ?T c = specific heat of water =4.18 J/g.C Q3 = (37 g)x(4.18 J/g·°C)[(25 °C - 0 °C)] Q3= (37 g)x(4.18 J/g·°C)x(25 °C) Q3 = 3866.5 J Heat required to raise the temperature of 0 °C water to 25 °C water = 3866.5 J the amount of Heat needed to convert 37 g of ice at -18C to water at 25C = Q1+Q2+Q3 =1391.94+12358+3866.5 = 17616.44 J =17.616KJ

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