Answer :
The energy required to heat 0.727 grams of water from -27.1 ℃ to 46.5 ℃ is calculated using the formula Q = mcΔT. Here, m = 0.000727 kg, c = 4.18 kJ/kg℃, and ΔT = 73.6℃. Substituting these values, we obtain the energy required as approximately 0.221 kJ.
Explanation:
To calculate the energy required to heat water, we can use the formula: Q = mcΔT, where Q is the heat energy, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.
The mass, m, is 0.727 grams or 0.727/1000 = 0.000727 kg, c = 4.18 kJ/kg℃ (this is the specific heat of water), and the change in temperature, ΔT, is 46.5℃ - (-27.1℃) = 73.6℃.
Plugging in our values we get: Q = (0.000727 kg) * (4.18 kJ/kg℃) * (73.6 ℃) = 0.221 kJ. Thus, the energy required to heat 0.727 grams of water from -27.1 ℃ to 46.5 ℃ is approximately 0.221 kJ.
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