How much heat in J must be added to 0.500 kg of water at room temperature (298 K) to raise its temperature to 65.0 ºC?
The specific heat of water is 4186 J/(kg∙ºC).
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The specific heat of water is 4186 J/(kg∙ºC).
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Verified answer
ΔQ = m*c*ΔT
ΔQ = 0.500kg * 4186J/kg·ºC * (65.0 - 25.0)ºC
ΔQ = 83 720 J ◄
to three significant digits.
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