Specific heat capacity

ANSWER: A
As the liquids are heated at the same rate, we just need to compare the energy required to raise the temperature of the liquids to their boiling points.
Energy = mcΔT = mc(boiling point - 20)
A. Energy = m(1000)(50 - 20) = 30000m
B. Energy = m(530)(60 - 20) = 21200m
C. Energy = m(850)(80 - 20) = 51000m
D. Energy = m(1710)(80 - 20) = 102600m
E. Energy = m(140)(360 - 20) = 47600m
Energy required for A is the least. Hence it will boil first.

2006-11-04 18:41:37 補充：
Sorry!!!! The answer should be BBBBBBBBBBB!!!!!!!!!!!!!!!What a careless mistake!!!!!!

2006-11-04 18:41:38 補充：
Sorry!!!! The answer should be BBBBBBBBBBB!!!!!!!!!!!!!!!What a careless mistake!!!!!!

The energy required to boil liquid A,

c=E/m△T

1000=E/(50-20)

1000*30=E

E=30000 J

The energy required to boil liquid B,

c=E/m△T

530=E/(60-20)

530*40=E

E=21200 J

The energy required to boil liquid C,

c=E/m△T

850=E/(80-20)

850*60=E

E=51000 J

The energy required to boil liquid D,

c=E/m△T

1710=E/(80-20)

1710*60=E

E=102600 J

The energy required to boil liquid E,

c=E/m△T

140=E/(360-20)

140*340=E

E=47600 J

∴Liquid B will boil first.

Notes:c=Specific Heat Capacity(J kg^-1 ℃^-1), E=Energy(Joule), m=mass(kg), △T=temperture change(℃)

Since the heating rate are the same, therefore the least energy required to raise the temperature of the liquid to its boiling point will boil first.

First let the mass of each liquid to be m kg.

For A, it requires to be raised by 30C to its boiling point,
the energy required is, E = m*C*( T2 - T1),
Ea = m*1000*( 50-20 ) = 30000m ( J )

Similiary,
Eb = m*530*( 60-20 ) = 21200m ( J )
Ec = m*850*( 80-20 ) = 51000m ( J )
Ed = m*1710*( 80-20 ) = 102600m ( J )
Ee = m*140*( 360-20) = 47600m ( J )

Eb is the smallest within the five value.
Therefore the least energy is required to raise the temperature of liquid B to its boiling point.
Therfeore liquid B will boil first.