B. Find the mass of the ice that was melted. Please answer Part B. Thank you so

Question

B. Find the mass of the ice that was melted.

Please answer Part B. Thank you so much in advance!

Safari File Edit View History Bookmarks Window Help webassign.net Phy 1010 Ch 11 Hom-W18 Guided Solutions and Help l EXAMPLE 11.6 Partial Melting Anki dmg GOAL Understand how to handle an incomplete phase change PROBLEM A 5.0o-kg block of ice at 0°C is added to an insulated container partially filled with 10.0 kg of water at 15.0°C. (a) Find the final temperature, neglecting the heat capacity of the container. (b) Find the mass of the ice that was melted STRATEGY Part (a) is tricky because the ice does not entirely melt in this example. When there is any doubt concerning whether there will be a complete phase change, some preliminary calculations are necessary. First, find the total energy required to melt the ice, met and then find water the maximum energy that can be delivered by the water above 0°C. If the energy delivered by the water is high enough, all the ice melts. If not, there will usually be a final mixture of ice and water at 0°C, unless the ice starts at a temperature far below 0°C, in which case all the liquid water freezes. Screen Shot 2018-03... 1.10 PM SOLUTION (A) Find the equilibrium temperature. Qmelt mce (5.00 kg)(3.33 x 105 J/kg) First, compute the amount of energy necessary to completely melt the ice: = 1.67 x 106 QwatermwatercAT Next, calculate the maximum energy that can be lost by the initial mass of liquid water without freezing it. (10.0 kg)(4190 -6.29 x 105 /kg . oC)(0°C-15.0°C) = This result is less than half the energy necessary to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point. (B) Compute the mass of ice melted 6.29 × 105 J = "tly = ""(3.33 x 105 J/kg) Set the total available energy equal to the heat of fusion of m grams of ice, mLy, and solve for m. m1.89 kg 27

Explanation / Answer

let m amount of ice melts

from principle of calorimetry

heat lost by hot body = heat gained by cold body

heat lost by water = heat gained by ice

-Mwater*Swater*(Tf - Twater) = m*L

-12*4190*(0-15) = m*3.33*10^5

mas of ice melted = 2.26 kg

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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