The amount of energy required for the vaporization of the ether:
Q = Q1 + Q2.
Q1 = C * m * (tc – tn), C is the specific heat of the ether (C = 2350 J / (K * kg)), m is the mass of the ether (m = 100 g = 0.1 kg), tc is the final temperature (tк = 35 ºС), tн – initial temperature (tн = 5 ºС).
Q2 = L * m, L – specific heat of vaporization of ether (L = 0.4 * 10 ^ 6 J / kg).
Q = Q1 + Q2 = C * m * (tк – tн) + L * m = 2350 * 0.1 * (35 – 5) + (0.4 * 10 ^ 6) * 0.1 = 7050 + 40000 = 47050 J = 47.05 kJ.
Answer: For the vaporization of ether, 47.05 kJ of heat is needed.
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