Question

If equal masses masses of paraffin wax mixed in container of negligible capacity, initial temperature of water is 80 °C and of paraffin is 20°C, then final temperature of mixture is ?​

Answer 1

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T

1

= 150

o

C

Final temperature of the metal, T

2

= 40

o

C

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm

3

Mass (M) of water at temperature T = 27

o

C:

150×1=150g

Fall in the temperature of the metal:

ΔT

m

=T

1

-T

2

=150−40=110

o

C

Specific heat of water, C

w

=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal, =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T

1

−T=40−27=13

o

C

Heat gained by the water and calorimeter system: =m

1

C

w

T=(M+m)C

w

T ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT

m

=(M+m)C

w

T

w

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg

−1

k

−1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Answer 2

The final temperature of mixture in the $60^{0}C$.

Explanation:

The final temperature = initial temperature-paraffin

$=80^{0}C-20^{0}C$

$=60^{0}C$

The final temperature of mixture $60^{0}$$C$.