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Answered

Calculate the energy needed to heat 10.5 g ice at -20.0 °C to liquid water at 75.0 °C. The heat of vaporization of water = 2257 J/g, the heat of fusion of water = 334 J/g, the specific heat capacity of water = 4.18 J/g·°C, and the specific heat capacity of ice = 2.06 J/g·°C.

Answer :

Drpelezogo

Answer:

= 7234.5Joules

Explanation:

Q = (mcΔT)ice + (mΔH)melting +(mcΔT)water

= (10.5g)(2.06J/g°C)[0°C-(-20°C)+(10.5g)(2257J/g)+(10.5g)(4.184J/g°C(75°C - 0°C)

= [(10.5)(2.06)(20) + (10.5)(2257) + (10.5)(4.184)(75)]J

= 432.6J + 3507J + 3294.9J

= 7234.5Joules

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