Question

At 1 atm, how much energy is required to heat 39.0 g of H2O(s) at –22.0 °C to H2O(g) at 169.0 °C? Helpful constants can be found here.

Answer 1

q = m * s * dT

m = mass of water

s = specific heat of water in different states, for ice = 2.108 oCJ/g, for water 4.187 0CJ/g and for vapour 1.996 0CJ/g

dT = change in teperature

q = m * S of ice * dT + n * dHf + m * S of water * dT + n * dHv + m * S of vapour * dT

dHf = enthapy of fusion and dHv = enthalpy of vapourisation

q = (39 * 2.108 * (0-(-22))) + ((39/18) * 6.02*10^3) + (39 * 4.187 * (100-0) + ((39/18) * 40.67*10^3) + 39 * 1.996 * (160-100)

q = 123.97 kJ