Adding Ice to Water Part A An insulated beaker with negligible mass contains liq

Question

Adding Ice to Water Part A An insulated beaker with negligible mass contains liquid water with a mass of 0.325 kg and a temperature of 62.3 C How much ice at a temperature of-10.9" C must be dropped into the water so that the final temperature of the system will be 24.0°C ? Take the specific heat of liquid water to be 4190 J/kg-K , the specific heat of ice to be 2100 J/kg-K, and the heat of fusion for water to be 3.34x105 J/kg Hints mice = 1.11 kg Submit My Answers Give Up Incorrect; Try Again; 8 attempts remaining Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures. Provide Feedback Continue

Explanation / Answer

Let the mass of ice be m

Then

Drop in energy of water = 0.325*4190*(62.3-24) = 52155.025 J

Now for ice.. gain in energy = m*(2100*10.9 + 3.34*10^5 + 4190*24) = 457450 m = 52155.025

m = 0.114 Kg (ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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