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Answer :
Final answer:
The amount of energy in Calories needed to completely boil 1.70 kg of water at 23â°C is approximately 140.89 Cal.
Explanation:
The specific heat of water is 4.184 J/g °C. To calculate the amount of energy required to completely boil 1.70 kg of water at 23â°C, we need to determine the temperature change from 23â°C to the boiling point, which is 100â°C. The total heat energy can be calculated using the formula: Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change.
Q = (1.70 kg)(4.184 J/g °C)(100â°C - 23â°C)
Q = (1.70 kg)(4.184 J/g °C)(77â°C)
Q = 589.04 kJ
Converting kJ to Calories, 1 kJ is equal to 0.239 Cal.
So, the amount of energy in Calories needed to completely boil 1.70 kg of water at 23â°C is approximately 140.89 Cal.
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