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how many joules are needed to convert 312.0g of ice at -35.0 degrees C to liquid water at 25.0 degrees C.

User Katrix
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1 Answer

1 vote

Heat needed : 159.33 kJ

Further explanation

The heat to change the phase can be formulated :

Q = mLf (melting/freezing)

Q = mLv (vaporization/condensation)

Lf=latent heat of fusion

Lv=latent heat of vaporization

The heat needed to raise the temperature

Q = m . c . Δt

The specific heat of ice is 2.09J/g°C; the specific heat of water is 4.182 J/g°C; the heat of fusion is 333.0 J/g.

1. heat to raise temperature from -35 °C to 0 °C


\tt Q=312* 2.09* (0-(-35)=22,822.8~J

2. phase change(ice to water)


\tt Q=312* 333=103,896~J

3. heat to raise temperature from 0 °C to 25 °C


\tt Q=312* 4.18* (25-0)=32,604~J

Total heat needed :

22,828.8 + 103,896+32,604=159,328.8 J=159.33 kJ

User Amb
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