Question

Calculate the energy needed to heat 14.6 g ice at -10.0 °C to liquid water at...

Calculate the energy needed to heat 14.6 g ice at -10.0 °C to liquid water at 70.0 °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.

Homework Answers

Answer #1

Ti = -10.0

Tf = 70.0

here

Cs = 2.06 J/g.oC

Heat required to convert solid from -10.0 to 0.0

Q1 = m*Cs*(Tf-Ti)

= 14.6 g * 2.06 J/g.oC *(0-(-10)) oC

= 300.76 J

Lf = 334.0

Heat required to convert solid to liquid at 0.0

Q2 = m*Lf

= 14.6 g *334.0 J/g

= 4876.4 J

Cl = 4.18 J/g.oC

Heat required to convert liquid from 0.0 to 70.0

Q3 = m*Cl*(Tf-Ti)

= 14.6 g * 4.18 J/g.oC *(70-0) oC

= 4271.96 J

Total heat required = Q1 + Q2 + Q3

= 300.76J + 4876.4J + 4271.96 J

= 9449 J

= 9.45*10^3 J (Taking into account significant figure)

Answer: 9.45*10^3 J

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