How much energy is required to change a 40 g ice cube from ice at - 14◦c to steam at 118◦c? the specific heat of ice is 2090 j/kg · ◦ c, the specific heat of water is 4186 j/kg · ◦ c, the specific heat of stream is 2010 j/kg · ◦ c, the heat of fusion is 3.33 * 105 j/kg, and the heat of vaporization is 2.26 * 106 j/kg. answer in units of j?

Question

Esteban

Physics

7 April, 19:22

How much energy is required to change a 40 g ice cube from ice at - 14◦c to steam at 118◦c? the specific heat of ice is 2090 j/kg · ◦ c, the specific heat of water is 4186 j/kg · ◦ c, the specific heat of stream is 2010 j/kg · ◦ c, the heat of fusion is 3.33 * 105 j/kg, and the heat of vaporization is 2.26 * 106 j/kg. answer in units of j?

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Darian Roach · Answer 1

From - 44°C to 0°C: Q1 = mCΔT = 40 * (2090/1000) * 44 = 3678.4 J

Melting at 0°C: Q2 = mHf = 40 * (3.33*10^5/1000) = 13320 J

Heating liquid water from 0°C to 100°C: Q3 = mCΔT = 40 * (4186/1000) * 100 = 16744 J

Vaporizing water to steam at 100°C: Q4 = mHvap = 40 * (2.26*10^6/100) = 90400 J

Heating steam from 100°C to 118°C: Q5 = mCΔT = 40 * (2010/1000) * 18 = 1447.2 J

Energy required, Q = Q1+Q2+Q3+Q4+Q5 = 3678.4+13320+16744+90400+1447.2 = 125589.6 J