You combine a 6.3kg block of ice at -10 degrees C with 1 kg of steam at 105 degr

Question

You combine a 6.3kg block of ice at -10 degrees C with 1 kg of steam at 105 degrees C in an isolated system. The heat of fusion for water is 333 kJ/kg, the heat of vaporization is 2256 kJ/kg, the specific heats of ice, water vapor, and liquid are 2.1 kJ/kg , 1.9kJ/kg, and 4.186 kJ/kg respectively. A) how much heat is lost by steam from cooling and condensing it into liquid water at 100 degrees C B)How much heat must be gained by the ice in order to heat and melt it into liquid water at 0 degrees C C) What is the state and temperature of the water after thermal equilibrium has been reached?

Explanation / Answer

First let's figure out how much heat energy will be liberated by the 175 grams of steam as it condenses into water at 100ºC, then cools to 0.0ºC. Water has a heat of vaporization of 2260 J/g, so: q = mL q = (175 g)(2260 J/g) = 396,000 J q = mcdT q = (175 g)(4.184 J/gºC)(100.0ºC) = 73,200 J So the total heat liberated by the steam as it changes phase then cools is 396,000 J + 73,200 J = 469,000 J Now let's figure the amount of heat liberated by the water as it cools from 20.0ºC to 0.0ºC. q = mcdT q = (100.0 g)(4.184 J/gºC)(20.0ºC) = 8370 J Okay, so the total heat liberated by the condensing and cooling steam AND by the cooling water is 469,000 J + 8370 J = 477,000 J Let's find out how much ice that would melt. The heat of fusion for ice is 334 J/g. q = mL m = q/L = (477,000 J)/(334 J/g) = 1430 grams, or 1.43 kg. So of the original 1.65 kg, about 0.22 kg would be left. To just melt the remainder of the ice, you need [(220 g)(334 J/g) =] 73,000 J of heat energy. To figure out how much steam at 100.0ºC is needed, we need to take into account the fact that some of the energy will be provided by the steam condensing, and some will be provided when the resulting water cools. Let q1 = the heat released by the steam condensing into liquid water. Let q2 = the heat released by the water cooling from 100.0ºC to 0.0ºC q1 = mL q2 = mcdT q1 + q2 = 73,000 J mL + mcdT = 73,000 J m(L + cdT) = 73,000 J m = (73,000 J) / (L + cdT) m = (73,000 J) / (2260 J/g + (4.184 J/gºC)(100.0ºC)) m = (73,000 J) / (2260 J/g + 418.4 J/g) m = (73,000 J) / (2678 J/g) m = 27 grams of steam. Good luck!

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.