document.write( "Question 377195: How much energy is required to heat 1.5 L of water at a temperature of 20 °C to the boiling point 100 ºC?
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Algebra.Com's Answer #268224 by rfadrogane(214)\"\" \"About 
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How much energy is required to heat 1.5 L of water at a temperature of 20 °C to the boiling point 100 ºC?
\n" ); document.write( " Sol'n:
\n" ); document.write( " Q = mCp(T2-T1)
\n" ); document.write( " m = mass density of water = 1 kg/L for Water, thus m = 1.5-kg
\n" ); document.write( " Cp = 4.186 kJ/kg-ºC, T2 = 100ºC & T1 = 20ºC
\n" ); document.write( " Then,
\n" ); document.write( " Q = 1.5(4.186)(100-20)
\n" ); document.write( " Q = 502.32 kJ ----answer\r
\n" ); document.write( "\n" ); document.write( "There is 1.5 L of water in a kettle on a cooking plate of 2.0 KW. It takes 7 minutes to heat the water from 20 C to the poiling point (100 C). How much electrical energy is required ? and what is the efficiency of the heating process?
\n" ); document.write( " Sol'n:
\n" ); document.write( " From above: Q = 502.32-kJ, energy needed to raise the temp. of water.
\n" ); document.write( " From: Q = (eff.)E
\n" ); document.write( " where: E - the energy needed by the kettle
\n" ); document.write( " eff.= efficiency of the kettle
\n" ); document.write( " E = Pt
\n" ); document.write( " = 2,0000 J/s x 7(60 s.)
\n" ); document.write( " E = 840-kJ ----answer
\n" ); document.write( "thus,
\n" ); document.write( " eff. = Q/E (100)
\n" ); document.write( " = 502.32/840 (100)
\n" ); document.write( " eff. = 59.8% ----answer\r
\n" ); document.write( "\n" ); document.write( "just send your question to my email:
\n" ); document.write( "rfadrogane@rocketmail.com
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