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Answer :
Final answer:
To heat 27.0g of water from 14.0°C to 28.7°C would require 1661.5836 joules or 397.0079 calories, which do not match any of the given options. The calculation involves the specific heat capacity of water and the temperature change.Thus optiond) Joules: 467J, Calories: 97.7cal is correct answer
Explanation:
To calculate the energy in joules required to heat water, we can use the formula Q = mcΔT, where 'm' is the mass of the water, 'c' is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given values: m = 27.0 g, c = 4.184 J/g °C, and ΔT = 28.7°C - 14.0°C = 14.7°C.
The calculation is as follows:
Q = (27.0 g) × (4.184 J/g °C) × (14.7°C) = 1661.5836 J
To convert joules to calories, we use the conversion factor 1 calorie = 4.184 joules.
Q = 1661.5836 J × (1 cal/4.184 J) = 397.0079 calories
Thus, the correct amount of energy required to heat the water is 1661.5836 J or 397.0079 calories, which is not exactly represented in options a) through d). Therefore, the correct amount of energy in joules and calories required to heat 27.0g water from 14.0°C to 28.7°C must be recalculated if the choices a) to d) are the only possible answers.
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