Calc 320g H2O Melting Ice @ 68°C

[lilkrazyrae]

How much ice could be melted by 320. grams of water at 68.0 degrees C?
Would I us mL + mcT=0 for the equation, and do the latent heat of ice for the first part and water for the second?

 

[Astronuc]

One must determine the energy (heat) from the mass of water decreasing from 68°C to 0°C. Then determine what mass of ice would be transformed to water at 0°C.

Assuming that L in your equation is latent heat, and by T you mean change in temperature, you are correct.

The heat from the liquid, Q, is just the change in energy in the liquid, i.e. Q = $\Delta\,H$ = $m\,c\,\Delta\,T$, where c is specific heat.

 

[lilkrazyrae]

So it would be m(333.7kJ/kg)+(.320kg)(4.186kJ/kg*K)(-68.0K)=.273kg of ice Is this correct?

 

[Astronuc]

Correct.

For further reference -

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html