A 18.8-g ice cube at -18.5oC is placed in 132 mL of water at 37.7oC. Specific he

Question

A 18.8-g ice cube at -18.5oC is placed in 132 mL of water at 37.7oC.
Specific heat of ice: 2090 J/kg K
Specific heat of water: 4186 J/kg K
Latent heat of fusion for water: 3.33 x 105 J/kg
(Ignore the heat capacity of the container)

Find the final temperature (in degrees in Celsius):

I keep getting different answers, and people keep giving me different answers and confusing me.


:/

Explanation / Answer

let final temp=t degree [t>0] so mi*si*18.5+Li*18.8/1000 +mi*sw*t=.132*sw*(37.7-t) 18.8/1000 * 18.5*2090 + 3.33*10^5*18.8*10^(-3) +18.8/1000 * 4186* t= .132*4186*(37.7-t) so 6987.302 + 78.7 t=20831.2 -552.55 t so t=21.9 degree

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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