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Calculated the energy to heat 32 grams of liqiud water from 47c to boiling?
Wiki User
∙ 16y ago
Q = mc (delta T) = 32.5 x 1 x (75-34) = 1332.5 cal
Wiki User
∙ 11y ago
Ethan10 Baller
To calculate the energy needed to heat the water, you can use the specific heat capacity of water, which is 4.184 J/g°C.
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Calculate the temperature change: boiling point of water (100°C) - initial temperature (47°C) = 53°C.
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Calculate the energy: Energy = mass (32 g) x specific heat capacity x temperature change.
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Substituting the values, we get Energy = 32 g x 4.184 J/g°C x 53°C = 7068.096 J or 7.07 kJ.
Wiki User
∙ 16y agoboiling =100c so need to heat by 100-47 = 53 degrees 53*32 =1696 calories
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