A 0.100 kg cube of ice is heated from -1200C to +1200C. Given that the specific

Question

A 0.100 kg cube of ice is heated from -1200C to +1200C. Given that the specific heat of ice, water, and steam are 2090 J/kg.0C, 4186 J/kg.0C, and 2010 J/kg.0C respectively and the latent heat of fusion and vaporization are 3.33 x 105 J/kg and 2.26 x 106 J/kg, find the energy transferred for the following processes: (a) Warming the ice from -1200C to the melting point (00C); (b) Melting the ice to become water; (c) Warming the resulting water to 1000C; (d) Vaporizing the water to become steam at 1000C (e) Heating the steam to 1200C.

Explanation / Answer

a)Q = .1 * 2090 * 1200 = 250800J b)Q = .1 * L = 33300J c)Q = .1*4186*1000 =418600J d)Q = .1 * 2.26*10^5 = 22600J e)Q = .1 * 2010*200 =40200J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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