Calculate the amount of energy needed to heat 346 g of liqui

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 ; vapH (H2O) = 40.8 kJ mol^-1

Please show your method. Thank you!

Solution

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 !