Question

How much energy is required to change a 52 g ice cube from ice at −9 ◦C to steam at 106◦C? The specific heat of ice is 2090 J/kg · ◦ C and of water 4186 J/kg · ◦ C. The latent heat of fusion of water is 3.33 × 105 J/kg, its latent heat of vaporization is 2.26 × 106 J/kg, and the specific heat of steam is 2010 J/kg · ◦ C. Answer in units of MJ.

Answer 1

mass of ice m = 52 gm = 0.052 kg

ice temperature Ti = 273 - 9 = 264 K

Water steam temperature Tst = 273 + 106 = 379 K

The specific heat of ice Ci = 2090 J/kg

The latent heat of fusion of water Li = 3.33 × 10⁵ J/kg

Now Change of heat dQ1, 52 g ice cube from ice at −9 ◦C is coverted into O ◦C ice is

dQ1 = mCi (T2 -T1) = 0.052x2090x(273 - 264) = 978 .12 J --------------------(1)

Now Change of heat dQ2, when 52 g ice cube is coverted into O ◦C water is

dQ2 = mLi = 0.52x3.33 × 10⁵ = 17316 J -------------------------(2)

specific heat of water Cw = 4186 J/kg

Now Change of heat dQ3, when 52 g water is coverted from O ◦C water to 106◦C is

dQ3 = mCw(Tf - Ti) = 0.052x4186 (379 - 273) = 23073.232 J -----------------------(3)

latent heat of vaporization Lv = 2.26 × 10⁶ J/kg

Now Change of heat dQ4, when 52 g water is coverted 106 ◦C water into 106◦C steam is

dQ4 =mLw = 0.052x2.26 × 10⁶ = 117520 J ---------------------------(4)

Now total energy (Q) is required to change a 52 g ice cube from ice at −9 ◦C to steam at 106◦C is

Q = dQ1 + dQ2 + dQ3 +dQ4 = 978 .12 + 17316 + 23073.232 + 117520 = 158887.352 J

Q = 0.159 MJ

dQ4

dQ3 =