Answer to Question #135008 in General Chemistry for Jayden

Question

A 529.78g sample of water at - 14.29 degrees * C is taken to a final temperature of 25.42 degrees * C How much energy was removed removed from the sample?

Answer 1

The given process includes three stages: 1) heating of ice from -14.29°C to 0°C; 2) melting of ice at 0°C; 3) heating of water from 0 to 25.42°C.

1) Q=cm∆T

Specific heat of ice c=2.108J/g°C,

∆T=0-(-14.29)=14.29°C,

Q=2.108×529.78×14.29=15958.73(J);

2)Q=cm

Heat of ice melting c=334 J/g,

Q=334×529.78=176946.52(J);

3)Q=cm∆T

Specific heat of water c=4.186 J/g°C,

∆T=25.42-0=25.42°C,

Q=4.186×529.78×25.42=56372.89(J)

Q_total=15958.73+176946.52+56372.89=249278.14(J)~249.28(kJ).