Calculate the amount of energy needed to heat 346 g of liquid water from 0°C to 182°C. Assume that the specific heat of water is 4.184 J g-1 °C-1 over the entire liquid range and that the specific heat of steam is 1.99 J g-1 °C-1 ; ΔvapH (H2O) = 40.8 kJ mol-1
Please show your method. Thank you!
Energy needed to heat water from 00C to 1000C = m*C*dT = 346*4.184*(100-0) = 144766.4 J
Energy needed to convert water to vapor at 1000C = n*L = (346/18)*40800 = 784266.6 J
Energy needed to heat vapor from 1000C to 1820C = m*C*dT = 346*1.99*(182-100) = 56460.28 J
So,
Total energy needed = 56460.28+784266.6+144766.4 = 985493.28 J = 985.49 kJ
Hope this helps !
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