The temperature of 1.00 mL of water is raised by 1.00 ^oC for every 4.184 joules of heat absorbed by the water.How many liters of water can be raised from 21.0 ^oC to 100.0 ^oC by the absorption of 8.88 kcal of heat generated by the combustion of natural gas? A)0)112 L B)2)13 L C)37.2 L D)168 L

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Quizplus · Accepted Answer

The correct answer is A.The amount of heat required to raise the temperature of 1.00 mL of water by 1.00 ^oC is 4.184 joules. Therefore, the amount of heat required to raise the temperature of 1.00 L of water by 1.00 ^oC is 4.184 kJ. The amount of heat required to raise the temperature of 1.00 L of water from 21.0 ^oC to 100.0 ^oC is 4.184 kJ/^oC * (100.0 ^oC - 21.0 ^oC) = 327.4 kJ. Therefore, the amount of heat required to raise the temperature of x L of water from 21.0 ^oC to 100.0 ^oC is 327.4 kJ * x L. The amount of heat generated by the combustion of 8.88 kcal of natural gas is 8.88 kcal * 4.184 kJ/kcal = 37.2 kJ. Therefore, the amount of water that can be raised from 21.0 ^oC to 100.0 ^oC by the absorption of 8.88 kcal of heat generated by the combustion of natural gas is 37.2 kJ / (327.4 kJ/L) = 0.112 L.

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